Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

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.

share|cite|improve this question
up vote 3 down vote accepted

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 =)

share|cite|improve this answer
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.

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.