Two methods to integrate? 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.
 A: Your first method is not correct. I suspect you are trying to use the following result which is not correct in general: $$\int\frac{1}{f(x)}dx = \log f(x)+C.$$
The correct form would be $$\int\frac{f'(x)}{f(x)}dx = \log f(x)+C,$$ which you cannot use in your example because the denominator of your integrand is not a linear function of $x$.
A: 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.
A: In method 1, you have a mistake. If $$u=2x^2$$ $$x=\frac{\sqrt{u}}{\sqrt{2}}$$ $$dx=\frac{1}{2 \sqrt{2} \sqrt{u}}$$ and $$\int \frac{x}{\sqrt{1 + 2x^2}}  dx=\frac{1}{4} \int \frac{du}{\sqrt{u+1}}=\frac{\sqrt{u+1}}{2}$$
A: In first you have:
$$\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
It should be:
$$\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}\int(1+u)^{-\frac{1}{2}}du=\frac{1}{4}\cdot 2 \sqrt{1+u}+C$$
A: The second answer is correct. The mistake in the first method is that your computation of the integral (after substitution) is wrong.
It holds:
$\int\frac{x}{\sqrt{1+2x^2}}dx=\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}\int(1+u)^{-\frac{1}{2}}u=\frac{2}{4} \sqrt{1+u} +C_2=\frac{1}{2}\sqrt{1+2x^2}+C_2$
A: Why not directly? Since
$$\int\frac1{\sqrt x}dx=\int x^{-1/2}dx=\frac{x^{1/2}}{1/2}+C=2\sqrt x+C$$
we get that for any differentiable (and positive) function $\;f\;$:
$$\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$
In our case, 
$$f(x)=1+2x^2\;,\;\;f'(x)=4x\implies\int\frac x{\sqrt{1+2x^2}}dx=\frac14\int\frac{(1+2x^2)'}{\sqrt{1+2x^2}}dx=$$
$$\frac14(2)\sqrt{1+2x^2}+C=\frac12\sqrt{1+2x^2}+C$$
