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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...

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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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  • $\begingroup$ So many people integrates a logarithmic derivative ($f'(x)/f(x)$) without substitution, but if we look at $f'(x)/(1+f(x)^2)$ the very same people first make a substitution - why is that? $\endgroup$ – AD. May 6 '11 at 5:03
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    $\begingroup$ @AD. If I'm teaching, I do the substitution in $\int (f'/f)\,dx$ as well. (In fact, I've discovered that skipping it tends to make a substantial number of students make the mistake of writing $\int(1/f(x))\,dx = \ln|f(x)|+C$ no matter what $f(x)$ is. Doing the substitution explicitly helps cut down on that particular mistake. $\endgroup$ – Arturo Magidin May 6 '11 at 5:05
  • $\begingroup$ You may have got a point there. However, if the student can not see $f'(x)/(1+f(x)^2)$ then it might be hard to see which substitution to do - also the more steps that is needed in a calculation the more mistakes seeps through too. $\endgroup$ – AD. May 6 '11 at 5:14
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    $\begingroup$ A quick way of factorizing $x^4+1$ is to write it as $x^4+1=(x^4+2x^2+1)-2x^2=(x^2+1)^2-(\sqrt{2}x)^2$, and then use $a^2-b^2=(a+b)(a-b)$ to get $[(x^2+1)+\sqrt{2}x][(x^2+1)-\sqrt{2}x]$. $\endgroup$ – Hans Lundmark May 6 '11 at 6:19
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    $\begingroup$ @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. $\endgroup$ – Hans Lundmark Jul 4 '11 at 19:37
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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?

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