integration using partial fraction can any body please tell me how to integrate the following expression:
$$\int\frac{x}{1+x^4}\,\mathrm {d}x$$ 
please help...
 A: 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? 
A: 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).
