Problem on indefinite integration $$ f(x) = \int\frac{x^2}{(1+x^2)(1 + \sqrt{1+x^2})}dx$$
where $f(0)=0$, find $f(1)$.
Despite many tries by taking $$\sqrt{1+x^2}=t $$  or 
$$ 1+\sqrt{1+x^2}$$ as $t$, I could not get it solved.
 A: Hint:
$$I = \int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})} \mathrm{d}x = \frac{x^2(\sqrt{1+x^2}-1)}{(1+x^2)(1+\sqrt{1+x^2})(\sqrt{1+x^2}-1)} \mathrm{d}x =\int \frac{\sqrt{x^2+1}-1}{1+x^2}\mathrm{d}x = \int \frac{1}{\sqrt{1+x^2}}-\frac{1}{1+x^2}\mathrm{d}x$$
Hope you can take it from here.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\mrm{f}\pars{x} & \equiv \int{x^{2} \over
\pars{1 + x^{2}}\pars{1 + \root{1 + x^{2}}}}\,\dd x
\end{align}

With Euler Substitution

$$
x = {1 - t^{2} \over 2t}\,\qquad t = \root{1 + x^{2}} - x
$$
$\ds{\mrm{f}\pars{x}}$ becomes
\begin{align}
\mrm{f}\pars{x} & =
\int\pars{-\,{1 \over t} + {2 \over t^{2} + 1}}\,\dd t =
-\ln\pars{t} + 2\arctan\pars{t}
\\[5mm] & =
2\arctan\pars{\root{1 + x^{2}} - x} - \ln\pars{\root{1 + x^{2}} - x} +
\pars{~\mbox{a constant}~}
\end{align}
A: HINT:
Let $x=\tan t$
$$\int\dfrac{x^2}{(1+x^2)({1+\sqrt{1+x^2)}}}dx=\int\dfrac{\tan^2t dt}{1+\sec t}$$
$$=\int\dfrac{1-\cos^2t}{\cos t(1+\cos t)}=\int\dfrac{1-\cos t}{\cos t}dt$$
See also: this
