Let I(x)= x^2(x ^2x+ x)(x x+1)^2dx.

Question

Let $I(x)=\int \frac{{x}^{2}(x{\sec }^{2}x+	an x)}{(x	an x+1{)}^{2}}dx$. If $I(0)=0$, then $I(\frac{\pi }{4})$ is equal to :

Accepted Answer

We have,

$\begin{gathered} I(x)= \int \frac{{x}^{2}(x{\sec }^{2}x+	an x)}{(x	an x+1{)}^{2}}dx \ \ = {x}^{2}\int \frac{x{\sec }^{2}x+	an x}{(x	an x+1{)}^{2}}dx \ \ \, -\int {\frac{d}{dx}({x}^{2})\int \frac{x{\sec }^{2}x+	an x}{(x	an x+1{)}^{2}}dx}dx $

$={x}^{2}(\frac{-1}{x	an x+1})+\int \frac{2x}{x	an x+1}dx$

Now, let

$ {I}_{1} =2\int \frac{x}{x	an x+1}dx \ \ =2\int \frac{x\cos x}{x\sin x+\cos x}dx \end{gathered}$

$\begin{gathered} x\sin x+\cos x=t \ \Rightarrow (x\cos x+\sin x-\sin x)dx=dt \ \Rightarrow x\cos xdx=dt \end{gathered}$

$\begin{gathered} 	herefore {I}_{1}=2\int \frac{dt}{t}=2\log t+c \ =2\log (x\sin x+\cos x)+c \end{gathered}$

$\begin{gathered} 	herefore I(x)=\frac{-{x}^{2}}{x	an x+1}+2\log (x\sin x+\cos x)+c \end{gathered}$

When, $x=0$, then

$\begin{gathered} I(0)=0+2\log (1)+c=0 \ \Rightarrow c=0 \ 	herefore I(x)=\frac{-{x}^{2}}{x	an x+1}+2\log (x\sin x+\cos x) \end{gathered}$

$\begin{gathered} 	herefore I(x)=\frac{-{x}^{2}}{x	an x+1}+2\log (x\sin x+\cos x) \ \Rightarrow I(\frac{\pi }{4})=\frac{-\frac{{\pi }^{2}}{16}}{\frac{\pi }{4}+1}+2\log (\frac{\pi }{4}	imes \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}) \ =\log (\frac{(\pi +4{)}^{2}}{32})-\frac{{\pi }^{2}}{4(\pi +4)} \end{gathered}$

Solutions · 1