Solving Improper Integrals of type 1

• August 9th 2010, 10:22 PM
dreamx20
Solving Improper Integrals of type 1
I ve been trying to solve this integral where it reads

int (-infinity to +infinity) x/((x^2+4)^3/2) so i wrote this as a limit where
lim A approaches -infinity (int from A to 0 x/((x^2+4)^3/2) + lim B approaches +infinity (int from 0 to B x/((x^2+4)^3/2). By subbing in u=x^2+4 i got the integral to equal -1/sqrt (X^2+4)+c and ended up with an answer of 0 when both integrals were added. is this correct i feel like this is wrong somehow
• August 9th 2010, 10:26 PM
yeKciM
you got correct result :D it's zero :D

$\displaystyle \int \frac {x\; dx }{\sqrt{(x^2+4)^{3}} } = -\frac {1}{\sqrt{(x^2+4)}} \big{|} _{-\infty} ^{+\infty} = 0$
• August 9th 2010, 10:29 PM
TheCoffeeMachine
That's correct. (Yes)