Integral of x^3 e^(x^2)/(x^2+1)^2

[bhatnv]

how would i go about finding the definite integral of this (x^3*e^(x^2))/(x^2+1)^2

 

[Mark44]

how would i go about finding the definite integral of this [PLAIN]http://www.texify.com/img/%5CLARGE%5C%21f%28x%29%3D%5Cfrac%7Bx%5E3e%5Ex%5E2%7D%7B%28x%5E2%2B1%29%5E2%7D.gif[/QUOTE]
Your integral doesn't show up.

 

[bhatnv]

there, originally had a image so it would be easier to read the equation, but i guess it got taken down

 

[Mark44]

Here are a couple of possibilities for you to explore:
1) u = x^2/(x^2 + 1)^2, dv = xe^(x^2)dx

2) u = x^2 e^(x^2), dv = x/(x^2 + 1)^2

I tried the first one, and it got pretty messy, so I don't think that's it.

 

[lgd0612]

By looking at
\frac{x^3 e^{x^2}}{(x^2+1)^2}
we can see its in the form of quotient rule because the (x^2+1)^2 at the bottom.
note that D_x [e^{x^2}] = 2x e^{x^2} and we can get the x^3 e^{x^2} term when we differentiate \frac{e^{x^2}}{x^2+1}


so the solution must be related to \frac {e^{x^2}}{x^2+1}

D_x [\frac {e^{x^2}}{x^2+1}] = \frac {e^{x^2}2x(x^2+1)-2xe^{x^2}}{(x^2+1)^2} = \frac {2x^3e^{x^2}}{(x^2+1)^2}