Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

please help...

share|improve this question

2 Answers 2

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).

share|improve this answer
    
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? –  AD. May 6 '11 at 5:03
1  
@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. –  Arturo Magidin May 6 '11 at 5:05
    
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. –  AD. May 6 '11 at 5:14
3  
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 $[(x^2+1)+\sqrt{2}x][(x^2+1)-\sqrt{2}x]$. –  Hans Lundmark May 6 '11 at 6:19
1  
@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. –  Hans Lundmark Jul 4 '11 at 19:37

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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.