Indefinite Integrals

We have

$ I=\int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x $

Let $\sec x+\tan x=t $

$\Rightarrow \sec x-\tan x=1 / t$.

$ \Rightarrow $ $\frac{1}{2}\left(t+\frac{1}{t}\right)=\sec x$

Now, $\left(\sec x \tan x+\sec ^2 x\right) d x=d t$

$ \Rightarrow $ $\sec x(\sec x+\tan x) d x=d t$

$ \Rightarrow $ $\sec x d x=\frac{d t}{t} $

$ I =\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right)}{t^{9 / 2}} \frac{d t}{t}$

$=\frac{1}{2} \int\left(t^{-9 / 2}+t^{-13 / 2}\right) d t $

$ =\frac{1}{2}\left[\frac{t^{-\frac{9}{2}+1}}{-\frac{9}{2}+1}+\frac{t^{-\frac{13}{2}+1}}{-\frac{13}{2}+1}\right] $

$=\frac{1}{2}\left[\frac{t^{-7 / 2}}{-\frac{7}{2}}+\frac{t^{-11 / 2}}{-\frac{11}{2}}\right]$

$=-\frac{1}{7} t^{-7 / 2}-\frac{1}{11} t^{-11 / 2}$

$=-\frac{1}{7} \frac{1}{t^{7 / 2}}-\frac{1}{11} \frac{1}{t^{11 / 2}}$

$=-\frac{1}{t^{11 / 2}}\left(\frac{1}{11}+\frac{t^2}{7}\right)$

$=-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$

Given,

$I = \int {{{{e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx} $

$J = \int {{{{e^{ - x}}} \over {{e^{ - 4x}} + {e^{ - 2x}} + 1}}dx = \int {{{{e^{3x}}} \over {{e^{4x}} + {e^{2x}} + 1}}dx} } $

$\therefore$ $J - I = \int {{{{e^x}({e^{2x}} - 1)} \over {{e^{4x}} + {e^{2x}} + 1}}dx} $

Let ${e^x} = t \Rightarrow {e^x}dx = dt$

$\therefore$ $J - I = \int {{{{t^2} - 1} \over {{t^4} + {t^2} + 1}}dt} $

$ = \int {{{1 - {1 \over {{t^2}}}} \over {{t^2} + 1 + {1 \over {{t^2}}}}}dt} $

Let, $t + {1 \over t} = u \Rightarrow \left( {1 - {1 \over {{t^2}}}} \right)dt = du$

$J - I = \int {{{du} \over {{u^2} - 1}} = {1 \over 2}\log \left| {{{u - 1} \over {u + 1}}} \right| + C} $

$ = {1 \over 2}\log \left| {{{{{{t^2} + 1} \over t} - 1} \over {{{{t^2} + 1} \over t} + 1}}} \right| + C$

$ = {1 \over 2}\log \left| {{{{e^{2x}} - {e^x} + 1} \over {{e^{2x}} + {e^x} + 1}}} \right| + C$

$\begin{aligned} & \text { Given, } f(x)=\frac{x}{\left(1+x^{n}\right)^{\frac{1}{n}}} \\ & \therefore f(f(x))=\frac{y}{\left(1+y^{n}\right)^{\frac{1}{n}}} \text {, where } y=\frac{x}{\left(1+x^{n}\right)^{\frac{1}{n}}} \\ & =\frac{x}{\left(1+2 x^{n}\right)^{\frac{1}{n}}} ~~......(\mathrm{i})\end{aligned}$

Similarly, $f(f(f(x)))=\frac{x}{\left(1+3 x^{n}\right)^{\frac{1}{n}}}$ and

$\begin{aligned} \frac{(f \circ f \circ f \circ \ldots f)(x)}{n \text { times }} & =g(x)=\frac{x}{\left(1+n x^{n}\right)^{\frac{1}{n}}}~~......(\mathrm{ii}) \\ \int x^{n-2} g(x) d x & =\int \frac{x^{n-1}}{\left(1+n x^{n}\right)^{1 / n}} d x \\ & =\frac{1}{n^{2}} \int n^{2}\left[x^{n-1}\left(1+n x^{n}\right)^{\frac{-1}{n}}\right] d x \\ & =\frac{1}{n^{2}} \cdot \frac{\left(1+n x^{n}\right)^{1-\frac{1}{n}}}{1-\frac{1}{n}}+k \\ & =\frac{\left(1+n x^{n}\right)^{1-\frac{1}{n}}}{n(n-1)}+k \end{aligned}$

$ \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} $

Taking $x^2$ from the expression under root

$ \begin{aligned} & =\int \frac{x^2-1}{x^5 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} d x \\ & =\int \frac{\frac{\left(x^2-1\right)}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} d x \\ & =\int \frac{\left(\frac{1}{x^3}-\frac{1}{x^5}\right)}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} d x \end{aligned} $

Substitute $2-\frac{2}{x^2}+\frac{1}{x^4}=u$

$ \begin{aligned} &\text { Differentiating both sides, we have }\\ &\begin{array}{r} \left(0+2 \cdot \frac{2}{x^3}-\frac{4}{x^5}\right) d x=d u \\ \Rightarrow \quad 4\left(\frac{1}{x^2}-\frac{1}{x^5}\right) d x=d u \end{array} \end{aligned} $

$ \begin{aligned} &\text { Now, above integral becomes after substitution. }\\ &\begin{aligned} \int \frac{d u}{4 \sqrt{u}} & =\frac{1}{4}\left(\frac{u^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right)+\mathrm{C} \\ & =\frac{1}{2} \sqrt{u}+\mathrm{C} \\ & =\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+\mathrm{C} \\ & =\frac{1}{2} \frac{\sqrt{2 x^4-2 x^2+1}}{x^2}+\mathrm{C} \\ & =\frac{\sqrt{2 x^4-2 x^2+1}}{2 x^2}+\mathrm{C} \end{aligned} \end{aligned} $