Unable to solve $ \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2} x + 1} dx $?

Question

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 ?

Answer 1

$x+\sqrt2 = \frac{1}{2}(2x+2\sqrt2) = \frac{1}{2}(2x+\sqrt2)+\frac{1}{2}\sqrt2$

Let $x^2+\sqrt2 \ x+1 = t \implies (2x+\sqrt2)dx = dt $

So, the integral is

$$I = \frac{1}{2}\int\frac{dt}{t}+\frac{1}{\sqrt2}\int\frac{dx}{(x^2+\sqrt 2 \ x+1)}$$

The first part gets evaluated into $\frac{1}{2}\ln t$. Convert the second part into $u^2+ a^2$ form which gets evaluated into $\frac{1}{a}\arctan(\frac{u}{a})$

Answer 2

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 $\arctan$.

Answer 3

Hint: We have $$x^2+\sqrt{2}x+1=\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}$$

Answer 4

$\displaystyle \frac{x+\sqrt{2}}{x^2+\sqrt2 x+1}=\frac{x+\sqrt{2}}{\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac12}$.

Let $\displaystyle x+\frac{\sqrt{2}}2=\frac{\sqrt2}2\tan\theta$.