TITLE: Two methods to integrate?
QUESTION [6 upvotes]: Are both methods to solve this equation correct? 
$$\int \frac{x}{\sqrt{1 + 2x^2}}  dx$$
Method One: 
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}}  du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{1}{4}\int\frac{du}{\sqrt{u}}$$
$$\frac{1}{2}\sqrt{u}+C$$
$$\frac{1}{2}\sqrt{1+2x^2}+C$$
I am confused why I get two different answers.

REPLY [3 votes]: Another one just for fun ;-)
Set $x=\frac{1}{\sqrt 2}\sinh(u)$ then,
$$\int^t\frac{x}{\sqrt{1+2x^2}}dx=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt 2 t)}\frac{\cosh(x)\sinh(u)}{\sqrt{1+\sinh^2(u)}}du=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt2 t)}\frac{\cosh(u)\sinh(u)}{\cosh u}du=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt2 t)}\sinh(u)du=\frac{1}{2}\cosh\left(\text{argsinh}(\sqrt 2 t)\right)+C=\frac{1}{2}\sqrt{2t^2+1}+C.$$ 
But it's not the shortest way to calculate it.