iCON Education - Integral Calculus

Q51 DPT Integration by Tricks MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

Evaluate $\int \frac{x}{\sqrt{x^2 + 9}} dx$.

A.

$\sqrt{x^2 + 9} + C$

B.

$2\sqrt{x^2 + 9} + C$

C.

$\frac{\sqrt{x^2 + 9}}{2} + C$

D.

$\frac{(x^2 + 9)^{3/2}}{3} + C$

Q52 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \frac{dx}{2x^2 + 7}$

A.

$\frac{1}{\sqrt{2}} \frac{1}{\sqrt{7}} \tan^{-1} \left( \frac{\sqrt{2}x}{\sqrt{7}} \right) + C$

B.

$\frac{1}{\sqrt{2}} \frac{1}{\sqrt{7}} \tan^{-1} \left( \frac{\sqrt{7}x}{\sqrt{2}} \right) + C$

C.

$\sqrt{2} \sqrt{7} \tan^{-1} \left( \frac{\sqrt{2}x}{\sqrt{7}} \right) + C$

D.

$\frac{1}{2} \frac{1}{7} \tan^{-1} \left( \frac{2x}{7} \right) + C$

Correct Answer: A

Explanation:

$\int \frac{dx}{2x^2 + 7}$
$= \int \frac{dx}{(\sqrt{2}x)^2 + (\sqrt{7})^2} \rightarrow \int \frac{dx}{x^2 + a^2}$
$= \frac{1}{\sqrt{2}} \frac{1}{\sqrt{7}} \tan^{-1} \left( \frac{\sqrt{2}x}{\sqrt{7}} \right) + C$

Q53 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \frac{dx}{x^2 - 16}$

A.

$\frac{1}{4} \ln \left| \frac{x-4}{x+4} \right| + C$

B.

$\frac{1}{8} \ln \left| \frac{x-4}{x+4} \right| + C$

C.

$\frac{1}{8} \ln \left| \frac{x+4}{x-4} \right| + C$

D.

$\frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + C$

Q54 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \frac{dx}{9 - x^2}$

A.

$\frac{1}{3} \ln \left| \frac{3+x}{3-x} \right| + C$

B.

$\frac{1}{6} \ln \left| \frac{3-x}{3+x} \right| + C$

C.

$\frac{1}{6} \ln \left| \frac{3+x}{3-x} \right| + C$

D.

$\frac{1}{6} \tan^{-1} \left( \frac{x}{3} \right) + C$

Q55 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \frac{dx}{\sqrt{9 - x^2}}$

A.

$\sin^{-1} \left( \frac{x}{9} \right) + C$

B.

$\sin^{-1} \left( \frac{x}{3} \right) + C$

C.

$\cos^{-1} \left( \frac{x}{3} \right) + C$

D.

$\frac{1}{3} \sin^{-1} \left( \frac{x}{3} \right) + C$

Q56 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \frac{dx}{\sqrt{x^2 + 9}}$

A.

$\ln \left| x + \sqrt{x^2 + 9} \right| + C$

B.

$\ln \left| x + \sqrt{x^2 + 3} \right| + C$

C.

$\frac{1}{3} \ln \left| x + \sqrt{x^2 + 9} \right| + C$

D.

$\sin^{-1} \left( \frac{x}{3} \right) + C$

Q57 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \frac{dx}{\sqrt{x^2 - 9}}$

A.

$\ln \left| x - \sqrt{x^2 - 9} \right| + C$

B.

$\frac{1}{3} \ln \left| x + \sqrt{x^2 - 9} \right| + C$

C.

$\ln \left| x + \sqrt{x^2 - 9} \right| + C$

D.

$\sin^{-1} \left( \frac{x}{3} \right) + C$

Q58 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \sqrt{9 - x^2} \, dx$

A.

$\frac{x}{2} \sqrt{9 - x^2} + \frac{9}{2} \sin^{-1} \left( \frac{x}{9} \right) + C$

B.

$\frac{x}{2} \sqrt{9 - x^2} + \frac{9}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

C.

$\frac{x}{2} \sqrt{9 - x^2} - \frac{9}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

D.

$\frac{x}{2} \sqrt{9 - x^2} + \frac{3}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

Q59 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \sqrt{x^2 + 9} \, dx$

A.

$\frac{x}{2} \sqrt{x^2 + 9} + \frac{9}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

B.

$\frac{x}{2} \sqrt{x^2 + 9} - \frac{9}{2} \ln \left| x + \sqrt{x^2 + 9} \right| + C$

C.

$\frac{x}{2} \sqrt{x^2 + 9} + \frac{9}{2} \ln \left| x + \sqrt{x^2 + 9} \right| + C$

D.

$\frac{x}{2} \sqrt{x^2 + 9} + \frac{3}{2} \ln \left| x + \sqrt{x^2 + 9} \right| + C$

Q60 DPT Formulae Practice MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Sep 2026

Concept

Evaluate $\int \sqrt{x^2 - 9} \, dx$

A.

$\frac{x}{2} \sqrt{x^2 - 9} + \frac{9}{2} \ln \left| x + \sqrt{x^2 - 9} \right| + C$

B.

$\frac{x}{2} \sqrt{x^2 - 9} - \frac{9}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

C.

$\frac{x}{2} \sqrt{x^2 - 9} - \frac{3}{2} \ln \left| x + \sqrt{x^2 - 9} \right| + C$

D.

$\frac{x}{2} \sqrt{x^2 - 9} - \frac{9}{2} \ln \left| x + \sqrt{x^2 - 9} \right| + C$

Q61 DPT Type -I ( Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 24 Sep 2026

Concept

Evaluate $\int \frac{dx}{x^2 - x - 6}$

A.

$\frac{1}{5} \ln \left| \frac{x+2}{x-3} \right| + C$

B.

$\frac{1}{5} \ln \left| \frac{x-3}{x+2} \right| + C$

C.

$\frac{1}{6} \ln \left| \frac{x-3}{x+2} \right| + C$

D.

$\ln \left| \frac{x-3}{x+2} \right| + C$

Q62 DPT Type -I ( Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 24 Sep 2026

Concept

Evaluate $\int \frac{dx}{x^2 - 7x + 10}$

A.

$\frac{1}{3} \ln \left| \frac{x-2}{x-5} \right| + C$

B.

$\frac{1}{3} \ln \left| \frac{x-5}{x-2} \right| + C$

C.

$\ln \left| \frac{x-5}{x-2} \right| + C$

D.

$\frac{1}{2} \ln \left| \frac{x-5}{x-2} \right| + C$

Q63 DPT Type -I ( Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 24 Sep 2026

Concept

Evaluate $\int \frac{\sin 2x \, dx}{(\sin^2 x - 1)(\sin^2 x + 3)}$

A.

$\frac{1}{4} \ln \left| \frac{\sin^2 x + 3}{\sin^2 x - 1} \right| + C$

B.

$\frac{1}{4} \ln \left| \frac{\sin^2 x - 1}{\sin^2 x + 3} \right| + C$

C.

$\frac{1}{2} \ln \left| \frac{\sin^2 x - 1}{\sin^2 x + 3} \right| + C$

D.

$\ln \left| \frac{\sin^2 x - 1}{\sin^2 x + 3} \right| + C$

Q64 DPT Type -I ( Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 24 Sep 2026

Concept

Evaluate $\int \frac{2e^{2x} \, dx}{(e^{2x} - 1)(e^{2x} - 9)}$

A.

$\frac{1}{8} \ln \left| \frac{e^{2x} - 1}{e^{2x} - 9} \right| + C$

B.

$\frac{1}{4} \ln \left| \frac{e^{2x} - 9}{e^{2x} - 1} \right| + C$

C.

$\frac{1}{8} \ln \left| \frac{e^{2x} - 9}{e^{2x} - 1} \right| + C$

D.

$\ln \left| \frac{e^{2x} - 9}{e^{2x} - 1} \right| + C$

Q65 DPT Type -I ( Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 24 Sep 2026

Concept

Evaluate $\int \frac{dx}{x^2 - 2x + 3}$

A.

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x+1}{\sqrt{2}}\right) + C$

B.

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x-1}{\sqrt{2}}\right) + C$

C.

$\frac{1}{2} \tan^{-1}\left(\frac{x-1}{2}\right) + C$

D.

$\sqrt{2} \tan^{-1}\left(\frac{x-1}{\sqrt{2}}\right) + C$

Q66 DPT Type -I ( Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 24 Sep 2026

Concept

Evaluate $\int \sqrt{3x - x^2} \, dx$

A.

$\frac{2x - 3}{2} \sqrt{3x - x^2} + \frac{9}{4} \sin^{-1}\left(\frac{2x - 3}{3}\right) + C$

B.

$\frac{2x - 3}{4} \sqrt{3x - x^2} - \frac{9}{8} \sin^{-1}\left(\frac{2x - 3}{3}\right) + C$

C.

$\frac{2x - 3}{4} \sqrt{3x - x^2} + \frac{9}{8} \sin^{-1}\left(\frac{2x - 3}{3}\right) + C$

D.

$\frac{2x + 3}{4} \sqrt{3x - x^2} + \frac{9}{8} \sin^{-1}\left(\frac{2x + 3}{3}\right) + C$

Correct Answer: C

Explanation:

$\int \sqrt{3x - x^2} \, dx = \int \sqrt{\left(\frac{3}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2} \, dx$
Using $\int \sqrt{a^2 - u^2} \, du$ with $u = x - \frac{3}{2}$ and $a = \frac{3}{2}$
$= \frac{x - 3/2}{2} \sqrt{3x - x^2} + \frac{(3/2)^2}{2} \sin^{-1}\left(\frac{x - 3/2}{3/2}\right) + C$
$= \frac{2x - 3}{4} \sqrt{3x - x^2} + \frac{9}{8} \sin^{-1}\left(\frac{2x - 3}{3}\right) + C$

Q67 DPT Type -I ( Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 24 Sep 2026

Concept

Evaluate $\int \frac{dx}{\sqrt{3x^2 + x - 2}}$

A.

$\sqrt{3} \ln\left| 6x + 1 + 2\sqrt{3} \sqrt{3x^2 + x - 2} \right| + C$

B.

$\frac{1}{\sqrt{3}} \ln\left| 6x - 1 + 2\sqrt{3} \sqrt{3x^2 + x - 2} \right| + C$

C.

$\frac{1}{\sqrt{3}} \ln\left| 6x + 1 + 2\sqrt{3} \sqrt{3x^2 + x - 2} \right| + C$

D.

$\frac{1}{\sqrt{3}} \sin^{-1}\left(\frac{6x + 1}{5}\right) + C$

Correct Answer: C

Explanation:

$\int \frac{dx}{\sqrt{3x^2 + x - 2}} = \frac{1}{\sqrt{3}} \int \frac{dx}{\sqrt{x^2 + x/3 - 2/3}}$
$= \frac{1}{\sqrt{3}} \int \frac{dx}{\sqrt{(x + 1/6)^2 - (1/6)^2 - 2/3}}$ where $-\frac{1}{36} - \frac{24}{36} = -\frac{25}{36}$
$= \frac{1}{\sqrt{3}} \int \frac{dx}{\sqrt{(x + 1/6)^2 - (5/6)^2}}$
Put $y = x + 1/6$: $= \frac{1}{\sqrt{3}} \int \frac{dy}{\sqrt{y^2 - a^2}}$ with $a = 5/6$
$= \frac{1}{\sqrt{3}} \ln\left| (x + 1/6) + \sqrt{x^2 + x/3 - 2/3} \right| + C = \frac{1}{\sqrt{3}} \ln\left| 6x + 1 + 2\sqrt{3} \sqrt{3x^2 + x - 2} \right| + C$

Q68 DPT Type -I ( Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 24 Sep 2026

Concept

Evaluate $\int \frac{dx}{3x^2 + 7x + 6}$

A.

$\frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{6x - 7}{\sqrt{23}}\right) + C$

B.

$\frac{1}{\sqrt{23}} \tan^{-1}\left(\frac{6x + 7}{\sqrt{23}}\right) + C$

C.

$\frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{6x + 7}{\sqrt{23}}\right) + C$

D.

$\frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{6x + 7}{23}\right) + C$

Correct Answer: C

Explanation:

$\int \frac{dx}{3x^2 + 7x + 6} = \frac{1}{3} \int \frac{dx}{x^2 + 7x/3 + 2}$
$= \frac{1}{3} \int \frac{dx}{(x + 7/6)^2 - (7/6)^2 + 2}$ where $2 - \frac{49}{36} = \frac{23}{36}$
$= \frac{1}{3} \int \frac{dx}{(x + 7/6)^2 + (\sqrt{23}/6)^2}$
$= \frac{1}{3} \cdot \frac{6}{\sqrt{23}} \tan^{-1}\left(\frac{x + 7/6}{\sqrt{23}/6}\right) + C$
$= \frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{6x + 7}{\sqrt{23}}\right) + C$

Q69 DPT Type-2 ( Linear/Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

Evaluate $\int \frac{2x+7}{3x+4} \, dx$

A.

$\frac{2}{3}x + \frac{13}{3}\ln|3x+4| + C$

B.

$\frac{2}{3}x + \frac{13}{9}\ln|3x+4| + C$

C.

$\frac{3}{2}x + \frac{13}{9}\ln|3x+4| + C$

D.

$\frac{2}{3}x - \frac{13}{9}\ln|3x+4| + C$

Q70 DPT Type-2 ( Linear/Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

Evaluate $\int \frac{3x-4}{5x+3} \, dx$

A.

$\frac{3}{5}x + \frac{29}{25}\ln|5x+3| + C$

B.

$\frac{5}{3}x - \frac{29}{25}\ln|5x+3| + C$

C.

$\frac{3}{5}x - \frac{29}{25}\ln|5x+3| + C$

D.

$\frac{3}{5}x - \frac{29}{5}\ln|5x+3| + C$

Q71 DPT Type-2 ( Linear/Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

Evaluate $\int \frac{x}{3x+7} \, dx$

A.

$\frac{1}{3}x + \frac{7}{9}\ln|3x+7| + C$

B.

$\frac{1}{3}x - \frac{7}{9}\ln|3x+7| + C$

C.

$\frac{1}{3}x - \frac{7}{3}\ln|3x+7| + C$

D.

$3x - \frac{7}{9}\ln|3x+7| + C$

Q72 DPT Type-2 ( Linear/Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

Evaluate $\int \frac{x}{\sqrt{2x+1}} \, dx$

A.

$\frac{1}{6}(2x+1)^{3/2} + \frac{1}{2}\sqrt{2x+1} + C$

B.

$\frac{1}{6}(2x+1)^{3/2} - \frac{1}{2}\sqrt{2x+1} + C$

C.

$\frac{1}{3}(2x+1)^{3/2} - \frac{1}{2}\sqrt{2x+1} + C$

D.

$\frac{1}{6}(2x+1)^{3/2} - \sqrt{2x+1} + C$

Correct Answer: B

Explanation:

$\int \frac{x}{\sqrt{2x+1}} \, dx = \frac{1}{2}\int \frac{2x+1}{\sqrt{2x+1}} \, dx + \left(0 - \frac{1}{2}\right)\int \frac{dx}{\sqrt{2x+1}}$
$= \frac{1}{2}\int \sqrt{2x+1} \, dx - \frac{1}{2}\int \frac{dx}{\sqrt{2x+1}}$
$= \frac{1}{2} \times \frac{2}{3} \times \frac{(2x+1)^{3/2}}{2} - \frac{1}{2} \times \frac{2\sqrt{2x+1}}{2} + C$
$= \frac{1}{6}(2x+1)^{3/2} - \frac{1}{2}\sqrt{2x+1} + C$

Q73 DPT Type-2 ( Linear/Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

Evaluate $\int \frac{x+2}{x^2-3x+1} \, dx$

A.

$\frac{1}{2}\ln|x^2-3x+1| - \frac{7}{2\sqrt{5}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

B.

$\ln|x^2-3x+1| + \frac{7}{2\sqrt{5}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

C.

$\frac{1}{2}\ln|x^2-3x+1| + \frac{7}{2\sqrt{5}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

D.

$\frac{1}{2}\ln|x^2-3x+1| + \frac{7}{\sqrt{5}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

Correct Answer: C

Explanation:

$x+2 = \frac{1}{2}(2x-3) + \left(2 - \left(-\frac{3}{2}\right)\right) = \frac{1}{2}(2x-3) + \frac{7}{2}$
$\int \frac{x+2}{x^2-3x+1} \, dx = \frac{1}{2}\int \frac{2x-3}{x^2-3x+1} \, dx + \frac{7}{2}\int \frac{dx}{x^2-3x+1}$
$= \frac{1}{2}\ln|x^2-3x+1| + \frac{7}{2}\int \frac{dx}{\left(x-\frac{3}{2}\right)^2 - \left(\frac{\sqrt{5}}{2}\right)^2}$
$= \frac{1}{2}\ln|x^2-3x+1| + \frac{7}{2} \times \frac{1}{2 \times \frac{\sqrt{5}}{2}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

Q74 DPT Type-2 ( Linear/Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

Evaluate $\int \frac{x-2}{x^2-4x+3} \, dx$

A.

$\log\sqrt{x^2-4x+3} + C$

B.

$x\log(x-3) - 2\log(x-2) + C$

C.

$\log[(x-3)(x-1)] + C$

D.

None of these

Q75 DPT Type-2 ( Linear/Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

Evaluate $\int \frac{2x-3}{\sqrt{4x^2+4x+2}} \, dx$

A.

$\frac{1}{2}\sqrt{4x^2+4x+2} + 2\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

B.

$\frac{1}{2}\sqrt{4x^2+4x+2} - 2\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

C.

$\sqrt{4x^2+4x+2} - 2\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

D.

$\frac{1}{2}\sqrt{4x^2+4x+2} - 4\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

Correct Answer: B

Explanation:

Let $2x-3 = A\frac{d}{dx}(4x^2+4x+2) + B = A(8x+4) + B$
$\Rightarrow 8A = 2$ and $4A + B = -3 \Rightarrow A = \frac{1}{4}$ and $B = -4$
$2x-3 = \frac{1}{4}(8x+4) - 4$
$I = \frac{1}{4}\int \frac{(8x+4)\,dx}{\sqrt{4x^2+4x+2}} - 4\int \frac{dx}{\sqrt{4x^2+4x+2}}$
Let $4x^2+4x+2 = t^2 \Rightarrow (8x+4)\,dx = 2t\,dt$
$I = \frac{1}{4}\int \frac{2t\,dt}{t} - 4\int \frac{dx}{\sqrt{(2x+1)^2+1}} = \frac{1}{2}t - 2\log|(2x+1)+\sqrt{4x^2+4x+2}| + C = \frac{1}{2}\sqrt{4x^2+4x+2} - 2\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

Q76 DPT Type-2 ( Linear/Quadratic) MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

Evaluate $\int \frac{x \, dx}{x^2+4x+5}$

A.

$\frac{1}{2}\log(x^2+4x+5) + 2\tan^{-1}(x) + C$

B.

$\frac{1}{2}\log(x^2+4x+5) - \tan^{-1}(x+2) + C$

C.

$\frac{1}{2}\log(x^2+4x+5) + \tan^{-1}(x+2) + C$

D.

$\frac{1}{2}\log(x^2+4x+5) - 2\tan^{-1}(x+2) + C$

Q77 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 28 Sep 2026

Concept

Evaluate $\int \frac{dx}{3\sin^2 x + 4\cos^2 x - 6\sin x \cos x}$

A.

$\frac{1}{\sqrt{3}}\tan^{-1}\left(\sqrt{3}(\tan x + 1)\right) + C$

B.

$\sqrt{3}\tan^{-1}\left(\sqrt{3}(\tan x - 1)\right) + C$

C.

$\frac{1}{\sqrt{3}}\tan^{-1}\left(\sqrt{3}(\tan x - 1)\right) + C$

D.

$\frac{1}{3}\tan^{-1}\left(\sqrt{3}(\tan x - 1)\right) + C$

Correct Answer: C

Explanation:

Divide by $\cos^2 x$
$I = \int \frac{\sec^2 x \, dx}{3\tan^2 x + 4 - 6\tan x}$
Let $\tan x = t$, $I = \int \frac{dt}{3t^2 - 6t + 4}$
Since $3t^2 - 6t + 4 = 3\left[t^2 - 2t + \frac{4}{3}\right]$
$= 3\left[(t-1)^2 - 1 + \frac{4}{3}\right] = 3\left[(t-1)^2 + \frac{1}{3}\right]$
Therefore $I = \int \frac{dt}{3\left((t-1)^2 + \frac{1}{3}\right)} = \frac{1}{3}\int \frac{dt}{(t-1)^2 + \frac{1}{3}}$
$= \frac{1}{3} \cdot \frac{1}{a}\tan^{-1}\left(\frac{t-1}{a}\right) + C$ (where $a = \frac{1}{\sqrt{3}}$)
$= \frac{1}{3} \cdot \sqrt{3}\tan^{-1}\left(\sqrt{3}(t-1)\right) + C$
$= \frac{1}{\sqrt{3}}\tan^{-1}\left(\sqrt{3}(\tan x - 1)\right) + C$

Q78 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{dx}{1 + 3 \sin^2 x}$

A.

$\frac{1}{3} \tan^{-1}(3 \tan x) + c$

B.

$\frac{1}{2} \tan^{-1}(2 \tan x) + c$

C.

$\tan^{-1}(\tan x) + c$

D.

None of these

Q79 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{dx}{2 - \sin^2 x}$

A.

$\frac{1}{2} \tan^{-1}\left(\frac{\tan x}{2}\right) + C$

B.

$\tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right) + C$

C.

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right) + C$

D.

None of these

Q80 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{dx}{1 + 3 \cos^2 x}$

A.

$\frac{1}{3} \tan^{-1}\left(\frac{\tan x}{3}\right) + C$

B.

$\frac{1}{2} \tan^{-1}\left(\frac{\tan x}{2}\right) + C$

C.

$2 \tan^{-1}\left(\frac{\tan x}{2}\right) + C$

D.

None of these

Q81 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{dx}{3 \sin^2 x + 4 \cos^2 x}$

A.

$\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3} \tan x}{2}\right) + C$

B.

$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3} \tan x}{2}\right) + C$

C.

$\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2 \tan x}{\sqrt{3}}\right) + C$

D.

None of these

Correct Answer: A

Explanation:

$\int \frac{dx}{3 \sin^2 x + 4 \cos^2 x} = \int \frac{\sec^2 x \, dx}{3 \tan^2 x + 4}$
$= \int \frac{dt}{3t^2 + 4}$ where $t = \tan x$
$= \frac{1}{3} \int \frac{dt}{t^2 + \left(2/\sqrt{3}\right)^2} = \frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{t}{2/\sqrt{3}}\right) + C = \frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3} \tan x}{2}\right) + C$

Q82 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{dx}{(2 \sin x + 3 \cos x)^2}$

A.

$\frac{1}{2(2 \tan x + 3)} + C$

B.

$-\frac{1}{2(2 \tan x + 3)} + C$

C.

$-\frac{1}{2 \tan x + 3} + C$

D.

None of these

Q83 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{dx}{3 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x}$

A.

$\frac{1}{\sqrt{11}} \tan^{-1}\left(\frac{3 \tan x + 2}{\sqrt{11}}\right) + C$

B.

$\frac{3}{\sqrt{11}} \tan^{-1}\left(\frac{3 \tan x + 2}{\sqrt{11}}\right) + C$

C.

$\frac{1}{\sqrt{11}} \tan^{-1}\left(\frac{\tan x + 2}{\sqrt{11}}\right) + C$

D.

None of these

Correct Answer: A

Explanation:

$I = \int \frac{\sec^2 x \, dx}{3 \tan^2 x + 4 \tan x + 5}$
Put $\tan x = t$, so that $\sec^2 x \, dx = dt$
$I = \int \frac{dt}{3t^2 + 4t + 5} = \frac{1}{3} \int \frac{dt}{t^2 + \frac{4}{3}t + \frac{5}{3}}$
$= \frac{1}{3} \int \frac{dt}{\left(t + \frac{2}{3}\right)^2 + \frac{11}{9}} = \frac{1}{3} \cdot \frac{3}{\sqrt{11}} \tan^{-1}\left(\frac{t + 2/3}{\sqrt{11}/3}\right) + C$
$= \frac{1}{\sqrt{11}} \tan^{-1}\left(\frac{3t + 2}{\sqrt{11}}\right) + C = \frac{1}{\sqrt{11}} \tan^{-1}\left(\frac{3 \tan x + 2}{\sqrt{11}}\right) + C$

Q84 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $I = \int \frac{dx}{\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x}$

A.

$\frac{1}{3} \tan^{-1}\left(\frac{\tan x - \frac{1}{\tan x}}{3}\right) + C$

B.

$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{\tan x - \frac{1}{\tan x}}{\sqrt{3}}\right) + C$

C.

$\sqrt{3} \tan^{-1}\left(\frac{\tan x - \frac{1}{\tan x}}{\sqrt{3}}\right) + C$

D.

None of these

Correct Answer: B

Explanation:

$I = \int \frac{\sec^4 x \, dx}{\tan^4 x + 1 + \tan^2 x}$
Put $\tan x = t$, $\sec^2 x \, dx = dt$
$I = \int \frac{(1 + t^2) \, dt}{t^4 + t^2 + 1} = \int \frac{\left(1 + \frac{1}{t^2}\right) dt}{t^2 + \frac{1}{t^2} + 1}$
$= \int \frac{\left(1 + \frac{1}{t^2}\right) dt}{\left(t - \frac{1}{t}\right)^2 + 3}$
Now, put $t - \frac{1}{t} = v \Rightarrow \left(1 + \frac{1}{t^2}\right) dt = dv$
$I = \int \frac{dv}{v^2 + (\sqrt{3})^2} = \frac{1}{\sqrt{3}} \tan^{-1} \frac{v}{\sqrt{3}} + C$
$= \frac{1}{\sqrt{3}} \tan^{-1} \left(\frac{\tan x - \frac{1}{\tan x}}{\sqrt{3}}\right) + C$

Q85 DPT Type-3 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $I = \int \frac{dx}{(\cos^2 x + 4 \sin^2 x)^2}$

A.

$\frac{5}{16} \tan^{-1}(2 \tan x) - \frac{3 \tan x}{8(1 + 4 \tan^2 x)} + C$

B.

$\frac{1}{16} \tan^{-1}(2 \tan x) + \frac{3 \tan x}{8(1 + 4 \tan^2 x)} + C$

C.

$\frac{5}{16} \tan^{-1}(2 \tan x) + \frac{3 \tan x}{8(1 + 4 \tan^2 x)} + C$

D.

None of these

Correct Answer: C

Explanation:

$I = \int \frac{dx}{\cos^4 x + 16 \sin^4 x + 8 \cos^2 x \sin^2 x}$
$= \int \frac{\sec^4 x \, dx}{1 + 16 \tan^4 x + 8 \tan^2 x}$
$= \int \frac{\sec^4 x \, dx}{1 + (2 \tan x)^4 + 2(2 \tan x)^2}$
Now, put $2 \tan x = \tan \theta \Rightarrow 2 \sec^2 x \, dx = \sec^2 \theta \, d\theta$
$= \frac{1}{2} \int \frac{\left(\frac{4 + \tan^2 \theta}{4}\right) \sec^2 \theta \, d\theta}{\sec^4 \theta}$
$= \frac{1}{8} \int (4 \cos^2 \theta + \sin^2 \theta) \, d\theta$
$= \frac{1}{8} \int (3 \cos^2 \theta + 1) \, d\theta = \frac{5}{16} \theta + \frac{3}{32} \sin 2\theta + C$
$= \frac{5}{16} \tan^{-1}(2 \tan x) + \frac{3 \tan x}{8(1 + 4 \tan^2 x)} + C$

Q86 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{1}{1 + \sin x + \cos x} \, dx$

A.

$\ln \left| \tan \frac{x}{2} - 1 \right| + C$

B.

$\ln \left| \tan \frac{x}{2} + 1 \right| + C$

C.

$\ln \left| 1 + \cot \frac{x}{2} \right| + C$

D.

None of these

Correct Answer: B

Explanation:

Let $I = \int \frac{1}{1 + \sin x + \cos x} \, dx$
$= \int \frac{1}{1 + \frac{2 \tan x/2}{1 + \tan^2 x/2} + \frac{1 - \tan^2 x/2}{1 + \tan^2 x/2}} \, dx$
$= \int \frac{1 + \tan^2 x/2}{1 + \tan^2 x/2 + 2 \tan x/2 + 1 - \tan^2 x/2} \, dx$
$= \int \frac{\sec^2 x/2}{2 + 2 \tan x/2} \, dx$
Putting $\tan \frac{x}{2} = t$ and $\frac{1}{2} \sec^2 \frac{x}{2} \, dx = dt$, we get
$I = \int \frac{1}{t + 1} \, dt = \ln|t + 1| + C = \ln \left| \tan \frac{x}{2} + 1 \right| + C$

Q87 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Integrate $I = \int \frac{dx}{4 + 5 \cos x}$

A.

$\frac{1}{6} \ln \left| \frac{3 + \tan \frac{x}{2}}{3 - \tan \frac{x}{2}} \right| + C$

B.

$\frac{1}{3} \ln \left| \frac{3 - \tan \frac{x}{2}}{3 + \tan \frac{x}{2}} \right| + C$

C.

$\frac{1}{3} \ln \left| \frac{3 + \tan \frac{x}{2}}{3 - \tan \frac{x}{2}} \right| + C$

D.

None of these

Correct Answer: C

Explanation:

put $\tan \frac{x}{2} = t \Rightarrow \cos x = \frac{1 - t^2}{1 + t^2}$, $dx = \frac{2 \, dt}{1 + t^2}$
$I = \int \frac{\frac{2 \, dt}{1 + t^2}}{4 + \frac{5(1 - t^2)}{1 + t^2}} = \int \frac{2 \, dt}{4(1 + t^2) + 5(1 - t^2)} = \int \frac{2 \, dt}{9 - t^2}$
$= 2 \cdot \frac{1}{6} \ln \left| \frac{3 + t}{3 - t} \right| + C = \frac{1}{3} \ln \left| \frac{3 + t}{3 - t} \right| + C$
$= \frac{1}{3} \ln \left| \frac{3 + \tan \frac{x}{2}}{3 - \tan \frac{x}{2}} \right| + C$

Q88 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Integrate $\int \frac{dx}{5 - 13 \sin x}$

A.

$\frac{1}{24} \ln \left| \frac{\tan \frac{x}{2} - 5}{5 \tan \frac{x}{2} - 1} \right| + C$

B.

$\frac{1}{12} \ln \left| \frac{\tan \frac{x}{2} - 5}{5 \tan \frac{x}{2} - 1} \right| + C$

C.

$\frac{1}{12} \ln \left| \frac{5 \tan \frac{x}{2} - 1}{\tan \frac{x}{2} - 5} \right| + C$

D.

None of these

Correct Answer: B

Explanation:

put $\tan \frac{x}{2} = t \Rightarrow \sin x = \frac{2t}{1 + t^2}$, $dx = \frac{2 \, dt}{1 + t^2}$
$I = \int \frac{\frac{2 \, dt}{1 + t^2}}{5 - \frac{26t}{1 + t^2}} = \int \frac{2 \, dt}{5(1 + t^2) - 26t} = \int \frac{2 \, dt}{5t^2 - 26t + 5} = \int \frac{2 \, dt}{(5t - 1)(t - 5)}$
$= \frac{1}{12} \int \left(\frac{1}{t - 5} - \frac{5}{5t - 1}\right) dt = \frac{1}{12} \ln \left| \frac{t - 5}{5t - 1} \right| + C$
$= \frac{1}{12} \ln \left| \frac{\tan \frac{x}{2} - 5}{5 \tan \frac{x}{2} - 1} \right| + C$

Q89 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{dx}{3 \cos x + 4 \sin x + 5}$

A.

$-\frac{1}{\tan \frac{x}{2} + 2} + C$

B.

$\frac{1}{\tan \frac{x}{2} + 2} + C$

C.

$-\frac{1}{2\left(\tan \frac{x}{2} + 2\right)} + C$

D.

None of these

Q90 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Integrate $\int \frac{dx}{13 + 3 \cos x + 4 \sin x}$

A.

$\frac{1}{5} \tan^{-1}\left(\frac{5 \tan \frac{x}{2} + 2}{6}\right) + C$

B.

$\frac{1}{6} \tan^{-1}\left(\frac{5 \tan \frac{x}{2} - 2}{6}\right) + C$

C.

$\frac{1}{6} \tan^{-1}\left(\frac{5 \tan \frac{x}{2} + 2}{6}\right) + C$

D.

None of these

Correct Answer: C

Explanation:

put $\tan \frac{x}{2} = t \Rightarrow \cos x = \frac{1 - t^2}{1 + t^2}$, $\sin x = \frac{2t}{1 + t^2}$, $dx = \frac{2 \, dt}{1 + t^2}$
$I = \int \frac{\frac{2 \, dt}{1 + t^2}}{13 + \frac{3(1 - t^2)}{1 + t^2} + \frac{8t}{1 + t^2}} = \int \frac{2 \, dt}{13(1 + t^2) + 3(1 - t^2) + 8t} = \int \frac{2 \, dt}{10t^2 + 8t + 16} = \int \frac{dt}{5t^2 + 4t + 8}$
$= \frac{1}{5} \int \frac{dt}{\left(t + \frac{2}{5}\right)^2 + \left(\frac{6}{5}\right)^2} = \frac{1}{5} \cdot \frac{5}{6} \tan^{-1}\left(\frac{5t + 2}{6}\right) + C = \frac{1}{6} \tan^{-1}\left(\frac{5 \tan \frac{x}{2} + 2}{6}\right) + C$

Q91 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{6 + 3\sin x + 14\cos x}{3 + 4\sin x + 5\cos x} dx$

A.

$2x - \ln|3 + 4\sin x + 5\cos x| + C$

B.

$x + 2\ln|3 + 4\sin x + 5\cos x| + C$

C.

$2x + \ln|3 + 4\sin x + 5\cos x| + C$

D.

$3x + \ln|3 + 4\sin x + 5\cos x| + C$

Correct Answer: C

Explanation:

$6 + 3\sin x + 14\cos x = A[3 + 4\sin x + 5\cos x] + B[4\cos x - 5\sin x] + C$
$\Rightarrow 6 + 3\sin x + 14\cos x = (3A + C) + (4A - 5B)\sin x + (5A + 4B)\cos x$
$6 = 3A + C$ ...(i)
$3 = 4A - 5B$ ...(ii)
$14 = 5A + 4B$ ...(iii)
$\Rightarrow A = 2, B = 1, C = 0$
$I = \int \frac{A D^r + B\left(\frac{d}{dx}D^r\right) + C}{D^r} dx$
$= \int A dx + B\int \frac{\frac{dD^r}{dx}}{D^r} dx + \int \frac{C}{D^r} dx$
$= 2x + \ln(3 + 4\sin x + 5\cos x) + c$

Q92 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{2\cos x + 3\sin x + 4}{3\cos x + 4\sin x + 5} dx$

A.

$\frac{18}{25}x + \frac{1}{25}\ln|3\cos x + 4\sin x + 5| + \frac{2}{25}\tan\left[\frac{x - \tan^{-1}(4/3)}{2}\right] + C$

B.

$\frac{18}{25}x - \frac{1}{25}\ln|3\cos x + 4\sin x + 5| - \frac{2}{25}\tan\left[\frac{x - \tan^{-1}(4/3)}{2}\right] + C$

C.

$\frac{18}{25}x - \frac{1}{25}\ln|3\cos x + 4\sin x + 5| + \frac{2}{25}\tan\left[\frac{x - \tan^{-1}(3/4)}{2}\right] + C$

D.

$\frac{18}{25}x - \frac{1}{25}\ln|3\cos x + 4\sin x + 5| + \frac{2}{25}\tan\left[\frac{x - \tan^{-1}(4/3)}{2}\right] + C$

Correct Answer: D

Explanation:

Let $2\cos x + 3\sin x + 4 = l(3\cos x + 4\sin x + 5) + m(-3\sin x + 4\cos x) + n$ ...(1)
Equating the coefficients of $\cos x$ and $\sin x$ on both sides of (1), and also equating the constant terms on both sides, we have
$2 = 3l + 4m$, $3 = 4l - 3m$, $4 = 5l + n$ ...(2)
Solving the system of equations (2), we have
$l = \frac{18}{25}$, $m = -\frac{1}{25}$, $n = \frac{2}{5}$ ...(3)
Substituting the expression for $2\cos x + 3\sin x + 4$ from (1), the given integrand and the values of $l, m, n$ from (3), we have
$\int \frac{2\cos x + 3\sin x + 4}{3\cos x + 4\sin x + 5} dx = \frac{18}{25}x - \frac{1}{25}\ln|3\cos x + 4\sin x + 5| + \frac{2}{5}\int \frac{dx}{3\cos x + 4\sin x + 5}$
$= \frac{18}{25}x - \frac{1}{25}\ln|3\cos x + 4\sin x + 5| + \frac{2}{25}\tan\left[\frac{x - (\tan^{-1}4/3)}{2}\right] + C$

Q93 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{2\sin x + 3\cos x}{3\sin x + 4\cos x} dx$

A.

$\frac{18}{25}x - \frac{1}{25}\ln|3\sin x + 4\cos x| + C$

B.

$\frac{18}{25}x + \frac{1}{25}\ln|3\sin x + 4\cos x| + C$

C.

$\frac{18}{25}x + \frac{1}{25}\ln|4\sin x + 3\cos x| + C$

D.

$\frac{1}{25}x + \frac{18}{25}\ln|3\sin x + 4\cos x| + C$

Correct Answer: B

Explanation:

Let $2\sin x + 3\cos x = l(3\sin x + 4\cos x) + m(3\cos x - 4\sin x)$
$= (3l - 4m)\sin x + (4l + 3m)\cos x$
Now comparing the coefficients of $\sin x$ and $\cos x$ of both sides, we get
$3l - 4m = 2$ and $4l + 3m = 3$,
$\Rightarrow l = \frac{18}{25}$, $m = \frac{1}{25}$.
$\therefore 2\sin x + 3\cos x = \frac{18}{25}(3\sin x + 4\cos x) + \frac{1}{25}(3\cos x - 4\sin x)$
$\therefore I = \frac{18}{25}\int dx + \frac{1}{25}\int \frac{3\cos x - 4\sin x}{3\sin x + 4\cos x} dx$
$= \frac{18}{25}x + \frac{1}{25}\ln|(3\sin x + 4\cos x)| + C$

Q94 DPT Type-4 and 5 MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Sep 2026

Concept

Evaluate $\int \frac{3\cos x + 2}{\sin x + 2\cos x + 3} dx$

A.

$\frac{6}{5}x - \frac{3}{5}\ln|\sin x + 2\cos x + 3| - \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

B.

$\frac{6}{5}x + \frac{3}{5}\ln|\sin x + 2\cos x + 3| + \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

C.

$\frac{6}{5}x + \frac{3}{5}\ln|\sin x + 2\cos x + 3| - \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

D.

$\frac{3}{5}x + \frac{6}{5}\ln|\sin x + 2\cos x + 3| - \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

Correct Answer: C

Explanation:

Let $I = \int \frac{3\cos x + 2}{\sin x + 2\cos x + 3} dx$.
Let $3\cos x + 2 = \lambda(\sin x + 2\cos x + 3) + \mu(\cos x - 2\sin x) + \nu$.
Comparing coefficients of $\sin x$, $\cos x$ and constant terms: $0 = \lambda - 2\mu$, $3 = 2\lambda + \mu$, $2 = 3\lambda + \nu$.
Solving: $\lambda = \frac{6}{5}$, $\mu = \frac{3}{5}$, $\nu = -\frac{8}{5}$.
$\therefore I = \frac{6}{5}\int dx + \frac{3}{5}\int \frac{\cos x - 2\sin x}{\sin x + 2\cos x + 3} dx - \frac{8}{5}\int \frac{dx}{\sin x + 2\cos x + 3}$
$= \frac{6}{5}x + \frac{3}{5}\ln|\sin x + 2\cos x + 3| - \frac{8}{5}\int \frac{2\,dt}{t^2 + 2t + 5}$ where $t = \tan(x/2)$
$= \frac{6}{5}x + \frac{3}{5}\ln|\sin x + 2\cos x + 3| - \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

Q95 DPT Other Types MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

Evaluate $\int \frac{4e^{x} + 6e^{-x}}{9e^{x} - 4e^{-x}} \, dx$.

A.

$\frac{19}{36}x + \frac{35}{36}\ln|9e^{x} - 4e^{-x}| + C$

B.

$-\frac{19}{36}x + \frac{35}{36}\ln|9e^{x} - 4e^{-x}| + C$

C.

$-\frac{19}{36}x - \frac{35}{36}\ln|9e^{x} - 4e^{-x}| + C$

D.

$\frac{35}{36}x - \frac{19}{36}\ln|9e^{x} - 4e^{-x}| + C$

Correct Answer: B

Explanation:

To evaluate I = $\int \frac{P(x)}{Q(x)} \, dx$, assume P(x) = AQ(x) + BQ'(x) ...(1) where A and B are constants.
Then I = $\int \frac{AQ(x) + BQ'(x)}{Q(x)} \, dx$ = $A \int dx + B \int \frac{Q'(x)}{Q(x)} \, dx$ = $Ax + B \ln|Q(x)| + C$.
From (1), by comparing coefficients of same type of terms, one gets constants A and B.
In the present problem, I = $\int \frac{4e^{x} + 6e^{-x}}{9e^{x} - 4e^{-x}} \, dx$.
Denominator Q(x) = $9e^{x} - 4e^{-x}$.
Numerator P(x) = $4e^{x} + 6e^{-x}$.
As Q'(x) = $9e^{x} + 4e^{-x}$, we take $4e^{x} + 6e^{-x} = A(9e^{x} - 4e^{-x}) + B(9e^{x} + 4e^{-x})$.
By comparing the coefficients of $e^{x}$ and $e^{-x}$, we get $4 = 9A + 9B$ and $6 = -4A + 4B$.
Therefore $A = -\frac{19}{36}$, $B = \frac{35}{36}$.
Therefore I = $\int \frac{A(9e^{x} - 4e^{-x}) + B(9e^{x} + 4e^{-x})}{9e^{x} - 4e^{-x}} \, dx$ = $A \int dx + B \int \frac{9e^{x} + 4e^{-x}}{9e^{x} - 4e^{-x}} \, dx$ = $Ax + B \ln|9e^{x} - 4e^{-x}| + C$ = $-\frac{19}{36}x + \frac{35}{36}\ln|9e^{x} - 4e^{-x}| + C$.

Q96 DPT Other Types MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

Evaluate $\int \frac{3e^{x} + 5e^{-x}}{4e^{x} - 5e^{-x}} \, dx$.

A.

$\frac{7}{8}\ln|4e^{x} - 5e^{-x}| + \frac{1}{8}x + C$

B.

$\frac{1}{8}\ln|4e^{x} - 5e^{-x}| - \frac{7}{8}x + C$

C.

$\frac{7}{8}\ln|4e^{x} - 5e^{-x}| - \frac{1}{8}x + C$

D.

$-\frac{7}{8}\ln|4e^{x} - 5e^{-x}| - \frac{1}{8}x + C$

Correct Answer: C

Explanation:

Let $(3e^{x} + 5e^{-x}) = A [4e^{x} + 5e^{-x}] + B [4e^{x} - 5e^{-x}]$
$\Rightarrow 3e^{x} + 5e^{-x} = e^{x}(4A + 4B) + e^{-x}[5A - 5B]$
$4A + 4B = 3 \Rightarrow A + B = \frac{3}{4}$ ...(i)
& $5A - 5B = 5 \Rightarrow A - B = 1$ ...(ii)
Adding (i) and (ii): $2A = \frac{7}{4} \Rightarrow A = \frac{7}{8}$; subtracting: $B = -\frac{1}{8}$.
Then I = $A \int \frac{d/dx(D^{r})}{D^{r}} \, dx + B \int dx = A \ln|D^{r}| + Bx + C$ = $\frac{7}{8}\ln|4e^{x} - 5e^{-x}| - \frac{1}{8}x + C$.

Q97 DPT Other Types MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

Evaluate $\int \tan 3x \, \tan 2x \, \tan x \, dx$.

A.

$\frac{1}{3}\ln|\sec 3x| - \frac{1}{2}\ln|\sec 2x| - \ln|\sec x| + C$

B.

$\frac{1}{3}\ln|\sec 3x| + \frac{1}{2}\ln|\sec 2x| + \ln|\sec x| + C$

C.

$\ln|\sec 3x| - \ln|\sec 2x| - \ln|\sec x| + C$

D.

$\frac{1}{3}\ln|\sec 3x| - \frac{1}{2}\ln|\sec 2x| + \ln|\sec x| + C$

Q98 DPT Other Types MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

Evaluate $\int \frac{\sin x}{\sin(x - \alpha)} \, dx$.

A.

$x \cos \alpha - \sin \alpha \ln|\sin(x - \alpha)| + C$

B.

$x \cos \alpha + \sin \alpha \ln|\sin(x - \alpha)| + C$

C.

$x \sin \alpha + \cos \alpha \ln|\sin(x - \alpha)| + C$

D.

$x \cos \alpha + \sin \alpha \ln|\cos(x - \alpha)| + C$

Q99 DPT Other Types MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

Evaluate $\int \frac{\sin 2x}{\sin(x + \frac{\pi}{4}) \sin(x - \frac{\pi}{4})} \, dx$.

A.

$\ln|\sin(x + \frac{\pi}{4})| - \ln|\sin(x - \frac{\pi}{4})| + C$

B.

$\ln|\cos(x + \frac{\pi}{4})| + \ln|\cos(x - \frac{\pi}{4})| + C$

C.

$\ln|\sin(x + \frac{\pi}{4})| + \ln|\sin(x - \frac{\pi}{4})| + C$

D.

$\frac{1}{2}\ln|\sin(x + \frac{\pi}{4})| + \frac{1}{2}\ln|\sin(x - \frac{\pi}{4})| + C$

Correct Answer: C

Explanation:

$\Rightarrow \int \frac{\sin((x - \frac{\pi}{4}) + (x + \frac{\pi}{4}))}{\sin(x + \frac{\pi}{4}) \sin(x - \frac{\pi}{4})} \, dx$
$\Rightarrow \int \frac{\sin(x - \frac{\pi}{4})\cos(x + \frac{\pi}{4}) + \cos(x - \frac{\pi}{4})\sin(x + \frac{\pi}{4})}{\sin(x + \frac{\pi}{4}) \sin(x - \frac{\pi}{4})} \, dx$
= $\int \cot(x + \frac{\pi}{4}) \, dx + \int \cot(x - \frac{\pi}{4}) \, dx$
$\Rightarrow \ln|\sin(x + \frac{\pi}{4})| + \ln|\sin(x - \frac{\pi}{4})| + C$.

Q100 DPT Other Types MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Sep 2026

Concept

Evaluate $\int \frac{dx}{\sin(x - a) \sin(x - b)}$.

A.

$\frac{1}{\sin(a - b)}[\ln|\sin(x - a)| + \ln|\sin(x - b)|] + C$

B.

$\frac{1}{\sin(a - b)}[\ln|\sin(x - a)| - \ln|\sin(x - b)|] + C$

C.

$\frac{1}{\sin(a - b)}[\ln|\sin(x - b)| - \ln|\sin(x - a)|] + C$

D.

$\frac{1}{\cos(a - b)}[\ln|\sin(x - a)| - \ln|\sin(x - b)|] + C$

Mathematics-II Chapters

Integral Calculus

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