How to find $\int{\frac{x}{\sqrt{x^2+1}}dx}$? I started simplifying  $$\int{\dfrac{x}{\sqrt{x^2+1}}dx}$$
but I always get this: 
$$\int{x(x^2+1)^{-1/2}dx}.$$
But I don't know how to follow by that way.

Substituting $x^2 + 1 = y$, hence $xdx = \frac{1}{2}dy$. Therefore, the integral becomes $\int \frac{1}{2\sqrt{y}}dy$ which should be simple to evaluate.
