Looks Simple Integral But it is fake

**parkhid** · June 12th 2010, 10:35 AM

HI , can anyone give me a way to solve :

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

**wonderboy1953** · June 12th 2010, 11:03 AM

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).

**chiph588@** · June 12th 2010, 11:41 AM

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.

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

**Ackbeet** · June 12th 2010, 01:10 PM

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

**chisigma** · June 12th 2010, 01:18 PM

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

$\chi$ $\sigma$

**Ackbeet** · June 12th 2010, 01:26 PM

Very true. My bad.

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