<s>[INST] Integrate the given expression to find $f(1)$ If $$f(x)=\int \frac {x^2}{(x^2+2)(1+\sqrt{x^2+1})}dx$$ and $f(0)=0$ , then find $f(1)$
I multiplied and divided by $\sqrt{x^2+1}-1$ but later got stuck at $\int \frac{\sqrt{x^2+1}}{x^2+2}.dx$ How to solve this integral. Could someone help me with this?
 [/INST] $$\int_{0}^{1}\frac{x^2\,dx}{(x^2+2)(1+\sqrt{1+x^2})}\stackrel{x\mapsto\sinh z}{=}\int_{0}^{\log(1+\sqrt{2})}\frac{\sinh^2(z)\cosh(z)\,dz}{(\sinh^2(z)+2)(1+\cosh z)} $$
equals, by the substitution $z=\log t$,
$$ \int_{1}^{1+\sqrt{2}}\frac{(1-t)^2 (1+t^2)}{t(1+6t^2+t^4)}\,dt $$
which can be computed by partial fraction decomposition. The roots of $t(1+6t^2+t^4)$ are given by $0$ and $\pm i(\sqrt{2}\pm 1)$ and $\int_{1}^{1+\sqrt{2}}\frac{dt}{t-\zeta}=\log\left(\frac{1+\sqrt{2}-\zeta}{1-\zeta}\right)$, hence by computing five residues and simplifying
$$ \int_{1}^{1+\sqrt{2}}\frac{(1-t)^2 (1+t^2)}{t(1+6t^2+t^4)}\,dt = \color{blue}{\log(1+\sqrt{2})-\tfrac{1}{\sqrt{2}}\left[\text{arctanh}\tfrac{1}{2}+\arctan\tfrac{1}{\sqrt{2}}\right]}.$$
</s>