# Thread: Improper Integral Help (Again)

1. ## Improper Integral Help (Again)

Doing this problem. I have the antiderivative after using partial fraction decomposition. What I do not understand is the answer. I thought that this was divergent, but it is convergent. When I look at the end product, I do not get how that converges.

Problem:
50. $\displaystyle \int^2_\infty \frac{dx}{(x+3)(x+1)^2}$

Now my question is, as lim R-> $\displaystyle \infty-$, does that cancel out the "ln" terms?

2. are you sure the integral is $\displaystyle \infty$ to $\displaystyle 2$

3. $\displaystyle log(x+3)-log(x+1)=log\frac{x+3}{x+1}=0$ as x tends to infinity

4. Amul you are right I put the domain backwards. Thank you Alex for that observation, totally forgot about that.

5. $\displaystyle \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+3)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}$

6. Originally Posted by Plato
$\displaystyle \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+1)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}$
Thank you for your help Plato .

7. Originally Posted by Plato
$\displaystyle \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+1)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}$
Am I missing something? When x= 0, the left side is $\displaystyle \frac{1}{3}$ and the right side is $\displaystyle \frac{1}{2}$. I don't believe those are equal!

8. Thanks. There is a typo.
Corrected.