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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?

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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

3 Answers 3

up vote 10 down vote accepted

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

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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 $k=-\frac14$.

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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

$$-\frac{1}{4(t+4)^2}+c$$

and making the inverse substitution we finally get

$$-\frac{1}{4(x^2+4)^2}+c$$

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