5
$\begingroup$

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 ?

$\endgroup$
3
  • 2
    $\begingroup$ Now we can try to open this topic. +1 for editing. $\endgroup$ – Michael Rozenberg Jun 16 '19 at 10:04
  • 1
    $\begingroup$ This or this are nearly equally good duplicate targets. There are undoubtedly better ones, but it is not my job to look for them. Anyway, this horse has been beaten to death. Downvotes to all trusted users who fail to realize this. Newbies are excused. $\endgroup$ – Jyrki Lahtonen Jun 23 '19 at 10:15
  • $\begingroup$ @arandomguy: you mentioned a "bigger" problem in your post. Does that serve simply a motivation to your question in the title or is that the real question you want to ask? $\endgroup$ – user9464 Jun 23 '19 at 21:04
1
$\begingroup$

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

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 $\frac{1}{a}\arctan(\frac{u}{a})$

$\endgroup$
3
$\begingroup$

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 $\arctan$.

$\endgroup$
1
  • 1
    $\begingroup$ (+1) nice ....... $\endgroup$ – E.H.E Jun 16 '19 at 4:24
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.