Solve the integral $\int \frac{dx}{\left(\sqrt{x^2+1}+x\right)^2}$ $$\int \frac{dx}{\left(\sqrt{x^2+1}+x\right)^2}$$
I think I need to use replace, but not sure: 
$$x=\frac{u^2-1}{2\cdot u}$$
$$dx=\left(1-\frac{u^2-1}{2u^2}\right)$$

Here, it might be a good idea to first simplify the integrand. We write
$$
\begin{aligned}
\frac{1}{\bigl(\sqrt{1+x^2}+x\bigr)^2}&=\frac{\bigl(\sqrt{1+x^2}-x\bigr)^2}{\bigl(\sqrt{1+x^2}+x\bigr)^2\bigl(\sqrt{1+x^2}-x\bigr)^2}\\
&=\bigl(\sqrt{1+x^2}-x\bigr)^2\\
&=1+2x^2-2x\sqrt{1+x^2}.
\end{aligned}
$$
Here, we have multiplied by the conjugate in the first step, simplified the denominator to 1 in the second step (conjugate rule twice, or squared if you want), and expanded the parenthesis in the last step. Thus,
$$
\int \frac{1}{\bigl(\sqrt{1+x^2}+x\bigr)^2}\,dx
=\int 1+2x^2-2x\sqrt{1+x^2}\,dx
=x+\frac{2}{3}x^3-\frac{2}{3}(1+x^2)^{3/2}+C.
$$
