# Improper Integral Help (Again)

• May 18th 2011, 05:05 AM
Warrenx
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. $\displaystyle \int^2_\infty \frac{dx}{(x+3)(x+1)^2}$

http://www4c.wolframalpha.com/Calcul...=25&w=457&h=40

Now my question is, as lim R-> $\displaystyle \infty-$, does that cancel out the "ln" terms?
• May 18th 2011, 05:24 AM
amul28
are you sure the integral is $\displaystyle \infty$ to $\displaystyle 2$
• May 18th 2011, 05:41 AM
alexmahone
$\displaystyle log(x+3)-log(x+1)=log\frac{x+3}{x+1}=0$ as x tends to infinity
• May 18th 2011, 01:17 PM
Warrenx
Amul you are right I put the domain backwards. Thank you Alex for that observation, totally forgot about that.
• May 18th 2011, 01:57 PM
Plato
$\displaystyle \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+3)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}$
• May 19th 2011, 02:33 AM
Warrenx
Quote:

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 :).
• May 19th 2011, 03:13 AM
HallsofIvy
Quote:

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!
• May 19th 2011, 03:34 AM
Plato
Thanks. There is a typo.
Corrected.