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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Im trying to solve the indefinite integral $$\int\frac{x}{(x^2+4)^3} \, \mathrm{d}x $$

I tried applying polynimial division and breaking to partial fractions but it didnt help...are there any other options?

share|cite|improve this question
Just as Mathmo123 answered, you face the situation where you have something looking as $\frac{u'}{u^3}$. So ... – Claude Leibovici Jul 6 '14 at 13:18
up vote 10 down vote accepted

Try the substitution $u=x^2+4$.

share|cite|improve this answer
simple and sufficent, thanks – Bak1139 Jul 6 '14 at 13:21

With questions like this, before jumping into anything more complicated like substitutions and partial fractions, sometimes its a good idea just to stop and think for a second, "what would I expect the solution to look like?" This won't always work, but when it does, it's a massive time saver.

Looking at this, rewriting it as $\int x(x^2+4)^{-3}dx$, I'd guess a solution of the form $$k(x^2 + 4)^{-2}$$ where $k$ is some constant that I need to find. I can then differentiate this expression to find out what $k$ should be:$$\frac{d}{dx}k(x^2 + 4)^{-2} = -4kx(x^2+4)^{-3} = x(x^2 + 4)^{-3}$$so $k=-\frac14$.

share|cite|improve this answer

Make the substitution $x^2=t$ to get $2xdx=dt$ and $dx=dt/(2x)$. Substituting in the integral you get $$ \int\frac{1}{2(t+4)^3} \, \mathrm{d}t$$

The indefinite integral is therefore easily obtained as


and making the inverse substitution we finally get


share|cite|improve this answer

Your Answer


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.