Evaluate $\int_0^\infty\frac{dx}{(x^3+(1+x^2)^{3/2}+x)(\sqrt{1+x^2} \tan^{-1} x+1)}$

Question

How to find this integral $$\int_0^\infty\frac{dx}{(x^3+(1+x^2)^{3/2}+x)(1+\sqrt{1+x^2} \arctan x)}$$ This one is very difficult for me because I missed my calc class twice and I didn't know about integrating inverse of trig function. Please help me. Thanks.

Answer 1

Rewrite the given integral as $$ \int_{x=0}^\infty\frac{dx}{\left(1+x^2\right)\left(\sqrt{1+x^2}+x\right)\left(\sqrt{1+x^2}\arctan x+1\right)}.\tag1 $$ Let $y=\tan x\;\Rightarrow\;dy=\sec^2x\ dx$ and $\sec x=\sqrt{1+x^2}$, the integral in $(1)$ turns out to be $$ \int_{y=0}^{\Large\frac\pi2}\frac{dy}{(\sec y+\tan y)(y\sec y+1)}.\tag2 $$ Multiply the numerator and denominator in $(2)$ by $\cos^2 y$ yields $$ \int_{y=0}^{\Large\frac\pi2}\frac{\cos^2 y\ dy}{(1+\sin y)(y+\cos y)}.\tag3 $$ The last step, let $u=y+\cos y\;\Rightarrow\;du=(1-\sin y)\ dy$, the integral in $(4)$ turns out to be $$ \int_{u=1}^{\Large\frac\pi2}\frac{1}{u}\ du=\large\color{blue}{\ln\left(\frac\pi2\right)}. $$

Answer 2

Let us denote the proposed integral by $I$. The substitution $u=\arctan x$ yields $$\eqalign{ I&=\int_0^{\pi/2}\frac{du}{(\tan u+\sec u)(u\sec u+1)}\cr &=\int_0^{\pi/2}\frac{\cos^2 u}{(1+\sin u)( u+\cos u)}du\cr &=\int_0^{\pi/2}\frac{1-\sin u}{ u+\cos u}du\cr &=\Big[\ln(u+\cos u)\big]_0^{\pi/2}=\ln\left(\frac{\pi}{2}\right) } $$ which is the desired result.$\qquad\square$