How to calculate this integral $\int \frac{x}{\sqrt{x^{2}+1}+x}\,dx$?

Question

Can somebody give mi idea, how to solve this integral?

$$\int \frac{x}{\sqrt{x^{2}+1}+x}\,dx$$

Answer 1

Hint: \begin{align} I&=\int \frac{x}{\sqrt{x^{2}+1}+x}\times\frac{\sqrt{x^{2}+1}-x}{\sqrt{x^{2}+1}-x}\,dx\\ &=\int \left(x\sqrt{x^{2}+1}-x^2\right)\,dx \end{align}

Answer 2

Expand by $\sqrt{x^2+1}-x$ to remove the root of the denominator. We then have $\int x\sqrt{x^2+1}-x^2 dx $. To solve $\int x\sqrt{x^2+1} dx$, we make the substitution $x^2+1 = u$. The other integral is easy to solve. Hope you can take it from here.