# Improper Integral Help

• May 18th 2011, 12:45 AM
Warrenx
Improper Integral Help
I am just starting on the improper integral section. I am having trouble with this problem because I cannot seem to break it down easily into something that does not have (x-5) in the denominator. If I try to do lim R -> 5+, I will still get an undefined answer. I am doing something wrong, but I do not know what.

12. $\displaystyle \int^6_5 \frac{dx}{(x-5)^{3/2}}$

I get the general antiderivative, but I cannot get rid of the denominator, and since the numerator is just 1, I cannot use L'Hopital's Rule. :(
• May 18th 2011, 01:07 AM
FernandoRevilla
Using the substitution $\displaystyle t=x-5$

$\displaystyle \displaystyle\int_5^6\dfrac{dx}{(x-5)^{3/2}}=\displaystyle\int_0^1\dfrac{dt}{t^{3/2}}=\displaystyle\lim_{ \epsilon \to 0^+}{\displaystyle\int_{\epsilon}^{ 1}\dfrac{dt}{t^{3/2}}}=\displaystyle\lim_{ \epsilon \to 0^+}\left[-\dfrac{2}{\sqrt{t}}\right]_{\epsilon}^1=\ldots$
• May 18th 2011, 01:20 AM
Warrenx
Thank you very much!
• May 18th 2011, 01:28 AM
FernandoRevilla
Quote:

Originally Posted by Warrenx
Thank you very much!

You are welcome!