3

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

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

please help...

Michael Medvinsky
  • 5,728
sadia abid
  • 121

2 Answers2

12

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).

Arturo Magidin
  • 398,050
  • 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 - Stop Putin - May 06 '11 at 05:03
  • 2
    @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 06 '11 at 05:05
  • 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 - Stop Putin - May 06 '11 at 05:14
  • @AD. The student will have a hard time seeing it in his mind's eye until he gets enough practice, just like he has a hard time seeing that $u=\cos x$ is the way to go with $\sin x e^{\cos x}$ until they actually do the substitution; noticing this sort of thing comes with practice and comfort. As to figuring out which substitutions to do, yes, they may have trouble seeing it, at least the first couple of times. – Arturo Magidin May 06 '11 at 05:17
  • 4
    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 06 '11 at 06:19
  • Arturo, should one of the denominators in your final equation be $x^2+\sqrt{2}x+1$? – yunone May 06 '11 at 06:23
  • @yunone: Yes, it should have. Thanks. – Arturo Magidin May 06 '11 at 14:54
  • @Hans: Did you use completing the square or something to see that conversion factor? – night owl Jul 04 '11 at 12:58
  • 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 04 '11 at 19:37
7

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?

yunone
  • 22,333