Integrate (ye^(x^2)) / (x^3) dx dy from sqrt(y) to 1. "y" Would be constant since it's a double integral integrated with respect to x first.

I used integration by parts and set u = e^(x^2) making du = 2xe^(x^2).

Setting dv = x^-3 makes v = (-1/2)x^(-2).

Thus uv - integral vdu = (e^(x^2)) / -(2(x^2)) + the integral (2xe^(x^2)) / (2x^2).
Which simplifies further to (e^(x^2)) / -(2(x^2)) + the integral e^(x^2) / x.

My problem comes from integrating e^(x^2) / x. When I use integration by parts I'm left with another integral that involves e^(x^2) which cant be integrated.

Question

Integrate (ye^(x^2)) / (x^3) dx dy from sqrt(y) to 1.   "y" Would be constant since it's a double integral integrated with respect to x first.

I used integration by parts and set u = e^(x^2) making du = 2xe^(x^2).

Setting dv = x^-3 makes v = (-1/2)x^(-2).

Thus uv - integral vdu = (e^(x^2)) / -(2(x^2)) + the integral (2xe^(x^2)) / (2x^2). 
Which simplifies further to (e^(x^2)) / -(2(x^2)) + the integral e^(x^2) / x.

My problem comes from integrating e^(x^2) / x. When I use integration by parts I'm left with another integral that involves e^(x^2) which cant be integrated.

anonymous · Answer

It seems to me like I'll be integrating the e^(x^2) an infinite amount of times. It appears there'll always be a (e^(x^2)) in the integrand of the integral v du. Unless I'm choosing my u and dv wrong.

anonymous · Answer

http://www.wolframalpha.com/input/?i=Integrate%5B%28Integrate%5B%28e%5Ex%5E2+y%29%2Fx%5E3%2C+x%5D%29%2C+%7By%2C+Sqrt%5By%5D%2C+1%7D%5D