# Thread:

1. ## Looks Simple Integral But it is fake

HI , can anyone give me a way to solve :

$<br /> \int \frac{x^2}{1+x^4}dx<br />$

2. ## Thought

I'm considering multiply both top and bottom by $x^{-2}$, then let $x = e^u$ which I believe converts to a familiar function (but you may also have to integrate by parts).

3. Originally Posted by parkhid
HI , can anyone give me a way to solve :

$<br /> \int \frac{x^2}{1+x^4}dx<br />$
Another way is to expand the integrand using partial fractions.

$\frac{x^2}{1+x^4}=\frac{1}{2\sqrt{2}}\cdot\frac{x} {x^2-\sqrt{2}x+1}-\frac{1}{2\sqrt{2}}\cdot\frac{x}{x^2+\sqrt{2}x+1}$

4. You can also use complex line integration. See here for a very similar sort of integral.

5. Originally Posted by Ackbeet
You can also use complex line integration. See here for a very similar sort of integral.
There is only a minor detail: the integral proposed by parkhid is indefinite and its solution is a family of functions... the integral of Your example is definite and its solution [if it exists...] is a real [or complex] number...

Kind regards

$\chi$ $\sigma$

6. Very true. My bad.

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