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can any body please tell me how to integrate the following expression:

$\int\frac{x}{1+x^4}\,\mathrm {d}x$

please help...

2 Answers 2

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Must you use partial fractions? Because taking $u=x^2$ gives $\int\frac{x}{1+x^4}\,dx = \frac{1}{2}\int\frac{1}{1+u^2}\,du$ and the last integral is immediate.

If you must use partial fractions, then your first step is to factor $1+x^4$ into a product of two irreducible quadratics. A simple way to do this is to go through the complex numbers. In the end, you get $x^4 + 1 = \left( x^2 - \sqrt{2}x + 1\right)\left(x^2+\sqrt{2}x + 1\right).$ So then you can do the partial fractions in the usual way, by expressing $\frac{x}{1+x^4}$ as $\frac{x}{1+x^4} = \frac{Ax+B}{x^2-\sqrt{2}x+1} + \frac{Cx+D}{x^2+\sqrt{2}x+1}$ for some constants $A$, $B$, $C$, and $D$.

But by far, substitution is the easier path (you'll have to do some substitution to solve the integrals you get after partial fractions anyway).

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    @night owl: Not really. You can factorize over the complex numbers by solving $z^4=1$ in polar coordinates, and then pair up the complex factors to get quadratic real factors, but after doing that (many years ago) I decided that it was a boring calculation, so I looked at the result and came up with this little shortcut which I memorized so that I wouldn't have to go through that pain again.2011-07-04
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I guess I should ask, is it imperative you use partial fractions? If so, I'll delete this answer since it's not quite what you're looking for.

Hint: Try a $u$-substitution, say $u=x^2$. Notice then that $du=2xdx$, so your integral now looks like $ \frac{1}{2}\int\frac{1}{1+u^2}du. $ Does this look familiar to something else you may have seen?