Indefinite Integration

Given that,

$ \begin{align*} I_m & =\int x^m \cos n x d x=g(x)-\frac{m(m-1)}{n^2} \\ I_{m-2} & =\int x^{m-2} \cos n x d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Solving integration by parts

$ =\cos n x \int x^{m-2} d x $

$ \begin{aligned} &\begin{aligned} & \left.-\int\left(\frac{d}{d x} \cos n x\right) \int x^{m-2} d x\right) d x \\ & =\cos n x\left[\frac{x^{m-1}}{m-1}\right]+n \int \sin n x \frac{x^{m-1}}{m-1} d x \\ & I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \\ & \int x^{m-1} \sin n x d x \end{aligned}\\ &\text { Again integrating by parts method, }\\ &\begin{aligned} & I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \\ & \left\{\sin n x \int x^{m-1} d x\right. \end{aligned} \end{aligned} $

$ \begin{aligned} & \left.-\int\left(\frac{d}{d x} \sin n x \int x^{m-1} d x\right) d x\right\} \\ = & \frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \\ & \left\{\frac{\sin n x \cdot x^m}{m}-\int n \cdot \cos n x \frac{x^m}{m} d x\right\} \end{aligned} $

$ \begin{aligned} & =\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \cdot \frac{1}{m} \\ & \left\{x^m \sin n x-n \int x^m \cos n x d x\right\} \\ & =\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} \end{aligned} $

$ \begin{aligned} & \left(x^m \sin n x-n I_m\right) \\ & I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} x^m \sin n x \end{aligned} $

$ -\frac{n^2}{m(m-1)} I_m $

$ \begin{aligned} &\begin{aligned} & \text { or } \frac{n^2}{m(m-1)} I_m=\frac{x^{m-1} \cos n x}{m-1} \\ & \quad+\frac{n}{m(m-1)} x^m \sin n x-I_{m-2} \\ & \text { or } I_m=\frac{m(m-1)}{n^2} \\ & {\left[\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} x^m \sin n x-I_{m-2}\right]} \\ & I_m=\frac{m}{n^2} x^{m-1} \cos n x+\frac{x^m \sin n x}{n} \\ & \quad-\frac{m(m-1)}{n^2} I_{m-2} \end{aligned}\\ &\text { On comparing with Eq. (i), we get }\\ &g(x)=\frac{x^m \sin n x}{n}+\frac{m}{n^2} x^{m-1} \cos n x \end{aligned} $

Given that, $I_n=\int \sec ^n x d x$

$ \begin{gathered} f(x)=5 I_6-4 I_4 \\ =5 \int \sec ^6 x d x-4 \int \sec ^4 x d x \\ =\int 5 \sec ^4 x \sec ^2 x d x-\int 4 \sec ^2 x \sec ^2 x d x \\ f(x)=\int\left\{5\left(1+\tan ^2 x\right)^2\right. \\ \left.\quad-4\left(1+\tan ^2 x\right)\right\} \sec ^2 x d x \end{gathered} $

Let $u=\tan x$

Differentiating w.r.t. ' $x$ ', we get

$ \begin{aligned} & \quad d u=\sec ^2 x d x \\ & \therefore f(x)=\int\left\{5\left(1+u^2\right)^2-4\left(1+u^2\right)\right\} d u \\ & =\int\left\{5\left(1+u^4+2 u^2\right)-4\left(1+u^2\right)\right\} d u \\ & =\int\left(5+5 u^4+10 u^2-4-4 u^2\right) d u \\ & =\int\left(5 u^4+6 u^2+1\right) d u=5 \frac{u^5}{5}+6 \frac{u^3}{3}+u \\ & f(x)=u^5+2 u^3+u \\ & f(x)=(\tan x)^5+2(\tan x)^3+\tan x \\ & \text { At } x=\frac{\pi}{4}, f\left(\frac{\pi}{4}\right)=\left(\tan \frac{\pi}{4}\right)^5+2\left(\tan \frac{\pi}{4}\right)^3 +\tan \frac{\pi}{4} \end{aligned} $

$ \begin{aligned} & =(1)^5+2(1)^3+(1) \\ f(\pi / 4) & =4 \end{aligned} $

$ \begin{aligned} &\, \text { We have, } \int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}} \\ \,\,\,\,\,\,& \quad=A \tan ^{-1} \sqrt{f(x)}+C \end{aligned} $

$ \text { } \begin{aligned}Let\,\, I & =\int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}} \\ & =\int \frac{x-1}{x(x+1) \sqrt{x+\frac{1}{x}+1}} d x \end{aligned} $

$ \begin{aligned} & \text { Put } x+\frac{1}{x}+1=t \text { and }\left(1-\frac{1}{x^2}\right) d x=d t \\ & =\int \frac{(x-1)}{x(x+1) \sqrt{t}} \times \frac{x^2}{\left(x^2-1\right)} d t \\ & =\int \frac{x}{(x+1)^2 \sqrt{t}} d t \\ & =\int \frac{d t}{\frac{x^2+2 x+1}{x} \sqrt{t}} \\ & =\int \frac{d t}{\left(1+x+\frac{1}{x}+1\right) \sqrt{t}} \\ & =\int \frac{d t}{(1+t) \sqrt{t}}=2 \tan ^{-1} \sqrt{t}+C \\ & =2 \tan ^{-1} \sqrt{x+\frac{1}{x}+1}+C \\ & \therefore \quad A=2 \Rightarrow f(x)=x+\frac{1}{x}+1 \\ & \,\,\,\,\,\, \quad A=2 \Rightarrow f(-1)=-1-1+1=-1 \\ & \therefore \quad(A, f(-1)=(2,-1) \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } f\left(\frac{2 x+3}{3 x+5}\right)=x+4 \\ & \text { Put, } \frac{2 x+3}{3 x+5}=y \\ & \Rightarrow \quad 2 x+3=3 x y+5 y \\ & \Rightarrow \quad x=\frac{5 y-3}{2-3 y} \\ & \therefore f(y)=\frac{5 y-3}{2-3 y}+4=\frac{5 y-3+8-12 y}{2-3 y} \\ & \Rightarrow f(y)=\frac{7 y-5}{3 x-2} \\ & \therefore f(x)=\frac{7 x-5}{3 x-2} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow f(y)=\frac{7 y-5}{3 x-2} \\ & \therefore f(x)=\frac{7 x-5}{3 x-2} \\ & \Rightarrow \int f(x) d x=\int \frac{7 x-5}{3 x-2} d x \end{aligned}\\ &\text { }\\ &\begin{aligned} Put\,\,3 x-2=t & \Rightarrow x=\frac{t+2}{3} \\ d x & =\frac{d t}{3}=\frac{1}{3} \int \frac{7\left(\frac{t+2}{3}\right)-5}{t} d t \\ & =\frac{1}{9} \int \frac{7 t-1}{t} d t \\ & =\frac{7}{9} \int d t-\frac{1}{3} \int \frac{d t}{t} \\ & =\frac{7}{9} t-\frac{1}{9} \log t+C \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{7}{9}(3 x-2)-\frac{1}{9} \log |3 x-2|+C \\ & =\frac{7}{3} x-\frac{1}{9} \log |3 x-2|+C \\ & \text { Here, } A=\frac{7}{3^{\prime}} B=-\frac{1}{9} \\ & \therefore 3 B-A=3\left(-\frac{1}{9}\right)-\frac{7}{3}=-\frac{8}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \int e^x\left(\frac{x^2-8 x+19}{(x-1)^5}\right) d x=\frac{e^x(l x+m)}{(x-1)^4}+C \\ & \text { Let } I=\int e^x\left(\frac{x^2-5 x+4-3 x+15}{(x-1)^5}\right) d x \\ & \quad I=\int e^x\left(\frac{(x-4)(x-1)}{(x-1)^5}-3 \frac{(x-5)}{(x-1)^5}\right) d x \\ & \quad I=\int e^x\left(\frac{x-4}{(x-1)^4}-\frac{3(x-5)}{(x-1)^5}\right) d x \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Here, } f(x)=\frac{x-4}{(x-1)^4} \\ & \Rightarrow \quad f(x)=\frac{-3(x-5)}{(x-1)^5} \end{aligned} $

$ \begin{aligned} & \therefore & I & \frac{e^x(x-4)}{(x-1)^4}+C \\ & \therefore & l & =1 m=-4 \\ & \therefore & 4 l+m & =4-4=0 \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}} \\ & \text { Put, } x-2=\frac{1}{t} d x=\frac{-1}{t^2} d t \Rightarrow x=2+\frac{1}{t} \\ & \begin{aligned} \therefore \quad I & =\int_{\frac{1}{t}}^{-1 / t^2} \frac{d t}{\sqrt{\left(2+\frac{1}{t}\right)^2-3\left(2+\frac{1}{t}\right)+5}} \\ I & =\int-\frac{d t}{\sqrt{3 t^2+t+1}} \\ & =\frac{-1}{\sqrt{3}} \int \frac{d t}{\left(t+\frac{1}{6}\right)^2+\frac{\sqrt{11}}{6}} \\ I & =\frac{-1}{\sqrt{3}} \sinh ^{-1}\left[\frac{6 t+1}{\sqrt{11}}\right] \\ I & =\frac{-1}{\sqrt{3}} \sinh ^{-1}\left[\frac{6+(x-2)}{\sqrt{11}(x-2)}\right]+C \end{aligned} \end{aligned} $

$ \begin{array}{r} {\left[t=\frac{1}{x-2}\right]} \\ I=\frac{-1}{\sqrt{3}} \sinh ^{-1}\left[\frac{x+4}{\sqrt{11}(x-2)}\right]+C \end{array} $

$ \begin{aligned} & \text { } \begin{array}{r} If\,\, \frac{2 x+1}{(x-1)^2\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2} +\frac{C x+D}{x^2+1} \end{array} \\ & \begin{array}{r} \Rightarrow 2 x+1=A(x-1)\left(x^2+1\right)+B\left(x^2+1\right) +(C x+D)(x-1)^2 \end{array} \\ & \Rightarrow 2 x+1=x^3(A+C) +x^2(-A+B-2 C+D) +x(A+C-2 D)+(-A+B+D) \end{aligned} $

On comparing coefficients of $x^3, x^2, x$ and constant terms both sides we get.

$ \begin{array}{r} A+C=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ -A+B-2 C+D=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ A+C-2 D=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\tag{iii} \end{array} $

$ \text { and } \quad-A+B+D=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

On solving Eqs. (i), (ii), (iii) and (iv), we get

$ \begin{array}{r} A=\frac{-1}{2}, B=\frac{3}{2}, C=\frac{1}{2} \text { and } D=-1 \\ \therefore A+B+C+D=\frac{-1}{2}+\frac{3}{2}+\frac{1}{2}-1=\frac{1}{2} \end{array} $

$ \begin{array}{r} \text { } \left.\int(\sqrt{1+\sin 2 x})+\sqrt{1-\sin 2 x}\right) d x \\ =\int\left(\sqrt{\sin ^2 x+\cos ^2 x+2 \sin x \cos x}\right. \\ \left.\quad+\sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x}\right) d x \\ =\int((\sin x+\cos x)+(\sin x-\cos x)) d x \\ \left(\because x \in\left(\frac{3 \pi}{4}, \pi\right)\right) \end{array} $

$ \begin{aligned} & =\int 2 \sin x d x=2 \int \sin x d x \\ & =2(-\cos x)+C=-2 \cos x+C \end{aligned} $

$ \begin{aligned} & \text {Let } I=\int x^2 \cdot \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x \\ & =x^2\left(-\frac{1}{(x \tan x+1)}\right) \\ & \quad-\int 2 x\left(-\frac{1}{(x \tan x+1)}\right) d x \\ & {\left[\text { since } \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2}=\int \frac{d t}{t^2}=-\frac{1}{t}\right.} \\ & \left.\quad=-\frac{1}{(x \tan x+1)}\right] \\ & =-\left(\frac{x^2}{x \tan x+1}\right)+\int \frac{2 x \cos x}{(x \sin x+\cos x)} d x \end{aligned} $

[putting $x \sin x+\cos x=u$

$ \begin{aligned} & \Rightarrow(x \cos x+\sin x-\sin x) d x=d u] \\ = & \frac{-x^2}{(x \tan x+1)}+2 \int \frac{d u}{u} \\ = & \frac{-x^2}{x \tan x+1}+2 \log |u|+C \\ = & \frac{-x^2}{x \tan x+1}+2 \log |x \sin x+\cos x|+C \\ \therefore & A=2 \text { and } B=-1 \text { and } f(x)=x^2 \\ \therefore & f(A+B)=f(2-1)=f(1)=(1)^2=1 \end{aligned} $

We have,

$ \begin{aligned} \int x^3(\log x)^2 d x & =x^4 \\ {\left[A(\log x)^2\right.} & +B(\log x)+C \log e]+K \end{aligned} $

Now, $\int x^3(\log x)^2 d x$

$ \begin{aligned} & =(\log x)^2 \int x^3 d x-\int\left(\frac{d}{d x}(\log x)^2 \int x^3 d x\right) d x \\ & \left.=(\log x)^2 \cdot \frac{x^4}{4}-\int(2 \log x) \cdot \frac{1}{x} \cdot \frac{x^4}{4}\right) d x \\ & =(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2} \int \log x \cdot x^3 d x \\ & =(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2} \\ & {\left[\log x \int x^3 d x-\int\left(\frac{d}{d x}\left(\log x \int x^3 d x\right) d x\right]\right.} \\ & =(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2} \\ & {\left[\log x\left(\frac{x^4}{4}\right)-\int \frac{1}{x} \cdot \frac{x^4}{4} d x\right]} \\ & =(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2} \\ & \quad\left[\log x\left(\frac{x^4}{4}\right)-\frac{1}{4} \cdot \int x^3 d x\right] \end{aligned} $

$ \begin{gathered} =(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2}\left[\log x\left(\frac{x^4}{4}\right)-\frac{1}{4} \cdot \frac{x^4}{4}\right]+K \\ =x^4\left[\frac{1}{4}(\log x)^2-\frac{1}{8} \log x+\frac{1}{32} \log e\right]+K \\ \therefore A=\frac{1}{4}, B=\frac{-1}{8} \text { and } C=\frac{1}{32} \\ \therefore A+B+C=\frac{1}{4}-\frac{1}{8}+\frac{1}{32} \\ =\frac{8-4+1}{32}=\frac{5}{32} \end{gathered} $

We have,

$ \begin{aligned} & \int \frac{9 x+15}{x^3-6 x-9} d x=A \log |g(x)| \\ & +B \log |f(x)|+C \end{aligned} $

Now,

$ \begin{aligned} & \frac{9 x+15}{x^3-6 x-9}=\frac{9 x+15}{(x-3)\left(x^2+3 x+3\right)} \\ & \begin{array}{r} \therefore \quad \frac{9 x+15}{(x-3)\left(x^2+3 x+3\right)} \\ =\frac{A}{x-3}+\frac{B x+C}{x^2+3 x+3} \end{array} \\ & \Rightarrow 9 x+15=A\left(x^2+3 x+3\right) \\ & \quad+(B x+c)(x-3) \\ & \Rightarrow 9 x+15=x^2(A+B)+x(3 A-3 B+C) \\ & \quad+(3 A-3 C) \end{aligned} $

On equating coefficients of $x^2, x$ and constant term, we get

$A+B=0,3 A-3 B+C=9$ and $3 A-3 C=15$

On solving above three equations we get, $A=2, B=-2$ and $C=-3$

$ \begin{aligned} & \therefore \int \frac{9 x+15}{(x-3)\left(x^2+3 x+3\right)} d x \\ & \quad=\int \frac{2}{x-3} d x+\int \frac{-2 x-3}{x^2+3 x+3} d x \\ & \quad=\int \frac{2}{x-3} d x-\int \frac{2 x+3}{x^2+3 x+3} d x \\ & \quad=2 \log |x-3|-\log \left|x^2+3 x+3\right|+C \\ & \therefore A=2, B=-1, g(x)=x-3 \text { and } \\ & f(x)=x^2+3 x+3 \\ & \therefore \frac{(A-B) g(4)}{f(-1)}=\frac{(2+1)(4-3)}{(-1)^2+3(-1)+3}=3 \end{aligned} $

We have,

$ \begin{aligned} & \frac{4 x^2+5 x^4+7}{\left(x^2+1\right)\left(x^4+x^2+1\right)}=\frac{A x+B}{x^2+1} \\ & +\frac{C x^3+D x^2+E x+F}{x^4+x^2+1} \\ & \Rightarrow 4 x^2+5 x^4+7=(A x+B)\left(x^4+x^2+1\right) \\ & +\left(C x^3+D x^2+E x+F\right)\left(x^2+1\right) \\ & \Rightarrow 5 x^4+4 x^2+7=A x^5+A x^3+A x \\ & +B x^4+B x^2+B+C x^5+C x^3+D x^4 \\ & +D x^2+E x^3+E x+F x^2+F \\ & \Rightarrow 5 x^4+4 x^2+7 \\ & =(A+C) x^5+(B+D) x^4+(A+C+E) x^3 \\ & +(B+D+F) \chi^2+(A+E) \chi+(B+F) \end{aligned} $

On comparing, we get

$ \begin{aligned} & A+C=0 \\ & B+D=5 \end{aligned} $

$ \begin{aligned} &\begin{aligned} A+C+E & =0 \\ B+D+F & =4 \\ A+E & =0 \\ B+F & =7 \end{aligned}\\ &\text { On solving, we get }\\ &\begin{aligned} & A=C=E=0, B=8 . D=-3, F=-1 \\ & \therefore B+2(D+F+E)-C \cdot A \\ & \quad \quad=8+2(-3-1+0)-0 \times 0 \\ & \quad=8-8=0 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} Let\,\,I & =\int \frac{y^2+\sqrt[3]{y^4}+\sqrt[6]{y^2}}{y\left(1+\sqrt[3]{y^2}\right)} d y \\ & =\int \frac{y^2+y^{4 / 3}+y^{1 / 3}}{y\left(1+y^{2 / 3}\right)} d y \\ & =\int \frac{y^{4 / 3}\left(y^{2 / 3}+1\right)+y^{1 / 3}}{y\left(1+y^{2 / 3}\right)} d y \end{aligned} $

            $ \begin{aligned} & =\int\left(y^{1 / 3}+\frac{y^{-2 / 3}}{1+y^{2 / 3}}\right) d y \\ & =\frac{y^{4 / 3}}{4 / 3}+\int \frac{y^{-2 / 3}}{1+\left(y^{1 / 3}\right)^2} d y \\ & =\frac{3}{4} \sqrt[3]{y^4}+3 \int \frac{d\left(y^{1 / 3}\right)}{1+\left(y^{1 / 3}\right)^2} \\ & =\frac{3}{4} \sqrt[3]{y^4}+3 \tan ^{-1} y^{1 / 3}+C \\ & =\frac{3}{4} \sqrt[3]{y^4}+3 \tan ^{-1} \sqrt[3]{y}+C \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{1}{1+k \cos x} d x \\ & \begin{aligned} =\int \frac{1}{1+k\left(\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}\right)} d x & \\ =\int \frac{\sec ^2 x / 2}{1+\tan ^2 \frac{x}{2}+k-k \tan ^2 \frac{x}{2}} d x \\ =\int \frac{\sec ^2 \frac{x}{2}}{(k+1)-(k-1) \tan ^2 \frac{x}{2}} d x \end{aligned} \end{aligned} $

Put, $\tan \frac{x}{2}=t \Rightarrow \sec ^2 \frac{x}{2} \cdot \frac{1}{2} d x=d t \Rightarrow$

$ \sec ^2 \frac{x}{2} d x=2 d t $

$ \begin{aligned} & \therefore \quad I=2 \int \frac{d t}{(k+1)-(k-1) t^2} \\ & =\frac{2}{k-1} \int \frac{d t}{\left(\sqrt{\frac{k+1}{k-1}}\right)^2-t^2} \\ & =\frac{2}{k-1} \cdot \frac{1}{2 \sqrt{\frac{k+1}{k-1}}} \log \left[\frac{\sqrt{\frac{k+1}{k-1}}+t}{\sqrt{\frac{k+1}{k-1}}-t}\right]+C \\ & =\frac{1}{\sqrt{k^2-1}} \log \left[\frac{\sqrt{k+1}+\sqrt{k-1} \tan \frac{x}{2}}{\sqrt{k+1}-\sqrt{k-1} \tan \frac{x}{2}}\right]+C \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int e^{-3 x}\left(x^2+\sin 4 x\right) d x \\ & =\int e^{-3 x} x^2 d x+\int e^{-3 x} \sin 4 x d x \\ & =x^2 \cdot \frac{e^{-3 x}}{-3}-\int-\frac{1}{3} e^{-3 x} \cdot 2 x d x+\frac{e^{-3 x}}{9+16} \\ & {[-3 \sin 4 x-4 \cos 4 x]} \end{aligned} $

$ \begin{aligned} & =-\frac{1}{3} x^2 e^{-3 x}+\frac{2}{3}\left[x \frac{e^{-3 x}}{-3}-\int-\frac{1}{3} e^{-3 x} \cdot 1 d x\right] \\ & +\frac{e^{-3 x}}{25}[-3 \sin 4 x-4 \cos 4 x]+C \\ & =-\frac{1}{3} x^2 e^{-3 x}-\frac{2}{9} x e^{-3 x}-\frac{2}{27} e^{-3 x} \\ & -\frac{3}{25} e^{-3 x} \sin 4 x-\frac{4}{25} e^{-3 x} \cos 4 x+C \end{aligned} $

$ \begin{array}{r} =-e^{-3 x}\left[\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x\right. \left.+\frac{4}{25} \cos 4 x\right]+C \end{array} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} Let\,\,I & =\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x \\ & =\int \frac{2 x^{12}+5 x^9}{\left[x^5\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)\right]^3} d x \\ & =\int \frac{2 x^{12}+5 x^9}{x^{15}\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} d x \\ & =\int \frac{2 x^{-3}+5 x^{-6}}{\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} d x \end{aligned}\\ &\begin{aligned} & \text { Put, } 1+\frac{1}{x^2}+\frac{1}{x^5}=t \\ & \begin{array}{l} \Rightarrow\left(-\frac{2}{x^3}-\frac{5}{x^6}\right) d x=d t \\ \therefore I=-\int \frac{d t}{t^3}=\frac{1}{2 t^2}+C \\ =\frac{1}{2\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^2}+C \\ =\int \frac{x^{10}}{2\left(x^5+x^3+1\right)^2}+C \\ \therefore m=10, l=2, r=2 \\ \therefore \quad \frac{m-l}{r}=\frac{10-2}{2}=4 \end{array} \end{aligned} \end{aligned} $

We have,

$I = \int {{{8{x^{43}} + 13{x^{38}}} \over {{{({x^{13}} + {x^5} + 1)}^4}}}dx} $

$ = \int {{{8{x^{43}} + 13{x^{38}}} \over {{x^{52}}\left( {1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}}} \right)}}dx} $

$ = \int {{{8\left( {{1 \over {{x^9}}}} \right) + 13\left( {{1 \over {{x^{14}}}}} \right)} \over {{{\left( {1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}}} \right)}^4}}}dx} $

Put, $1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}} = t$

$ \Rightarrow \left( {{{ - 8} \over {{x^9}}} - {{13} \over {{x^{14}}}}} \right)dx = dt$

$\therefore$ $I = - \int {{{dt} \over {{t^4}}} = {1 \over {3{t^3}}} + C} $

$\eqalign{ & \Rightarrow I = {1 \over {3{{\left( {1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}}} \right)}^3}}} + C = {{{x^{39}}} \over {3{{({x^{13}} + {x^5} + 1)}^3}}} + C \cr & \cr} $