Unable to solve $ \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2} x + 1} dx $? This comes from a bigger problem :- 
$$ \text{Evaluate } \int\frac{dx}{1+x^4} $$
After making $ \int \frac {dx}{1+x^4} = \frac{dx}{(1+x^2)^2 - (\sqrt{2}x)^2} $ and then applying partial fraction method, I got :-
$$ \int \frac{dx}{1 + x^4} =\frac{1}{2 \sqrt{2}} \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2}x + 1} dx - \frac{1}{2 \sqrt{2}} \int \frac{x - \sqrt{2}}{x^2 - \sqrt{2}x + 1} dx $$
Now, to the first integral, I tried making a u-substitution:-
$$ \text{Let }x^2 + \sqrt{2}x + 1 = u \\
\frac{du}{dx} = 2x + \sqrt{2} \\
\implies du = (2x + \sqrt{2}) dx \\ $$
As you can see, it is not the same as the numerator, which is $$ (x + \sqrt{2}) dx $$
Any hints on how to proceed ?

The hint:
$$\frac{x+\sqrt2}{x^2+\sqrt2x+1}=\frac{x+\frac{1}{\sqrt2}+\frac{1}{\sqrt2}}{x^2+\sqrt2x+1}$$ and use $\ln$ and $\arctan$. 
