Evaluate $\int_0^\infty\frac{dx}{(x^3+(1+x^2)^{3/2}+x)(\sqrt{1+x^2} \tan^{-1} x+1)}$ 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.
 A: 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
$$
\int_{u=1}^{\Large\frac\pi2}\frac{1}{u}\ du=\large\color{blue}{\ln\left(\frac\pi2\right)}.
$$
A: 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.$\qquad\square$
