# How would one integrate $\frac{3}{x^2(x+1)}$?

How would one integrate $\frac{3}{x^2(x+1)}$? I tried to use U-substitution. I made $u = x + 1$, so $du = dx$ and $x = u - 1$. However, I don't think this made the problem easier.

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As MarkBennet said, the key here is partial fractions. I will show two methods of doing this below. Both contains some "trickery", but after a few times these tricks will come natuarly.

The easiest is to see that

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

and now the same technique can be used to split $1/x(x+1)$, I will leave this as an excercise for you. A even more trickister approach is shown below. Note

$$\frac{1}{x^2(x+1)} = \frac{1-x^2+x^2}{x^2(x+1)} = \frac{(x+1)(1-x)}{x^2(x+1)}+\frac{x^2}{x^2(x+1)} = \frac{1}{x^2} - \frac{1}{x} + \frac{1}{x+1}$$. And I guess you are able to integrate the last terms by yourself =)

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In the second last formula is a sign error. –  Alex Oct 12 '12 at 7:43

The standard technique is to use Partial Fractions in the form $\cfrac A {x^2} + \cfrac B x + \cfrac C {(x+1)}$ - in this case$$\frac 1 {x^2(x+1)} = \cfrac 1 {x^2} - \cfrac 1 x + \cfrac 1 {(x+1)}$$

There is plenty of literature on the method for determining the right form of decomposition and efficient ways of identifying the coefficients.

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