# improper integral

• Feb 26th 2010, 03:15 PM
cdlegendary
improper integral
http://hw.math.ucsb.edu/webwork/math...220467img1.gif

so:

http://hw.math.ucsb.edu/webwork/math...220467img3.gif

I know you're supposed to take the limit of the integrals, but I'm having trouble getting started.
Without the limits, I found the indefinite integral to be 12arctan(sqrt(x)) +C
• Feb 26th 2010, 03:25 PM
Krizalid
well it converges, but you don't need to split it into two integrals, just put $x=t^2.$
• Feb 26th 2010, 03:31 PM
cdlegendary
Quote:

Originally Posted by Krizalid
well it converges, but you don't need to split it into two integrals, just put $x=t^2.$

I'm still confused. how would that work? (Headbang)
• Feb 26th 2010, 03:32 PM
Krizalid
the substitution? just apply it, can you show what do you get?
• Feb 26th 2010, 03:42 PM
cdlegendary
just subbing it in would just give $\int\frac{6}{t(1+t^2)}$ would it not?
• Feb 26th 2010, 03:43 PM
Krizalid
no it's wrong, try again, the integrand has de arctangent form.
• Feb 26th 2010, 03:51 PM
cdlegendary
so the integral would be $12arctan(t)+C$ which is equal to $12arctan(\sqrt(x))+C$ right?

but after finding the integral, what do you do next to find the definite value from 0 to infinity?