How do you integrate #int x^3 e^(x^2 ) dx # using integration by parts?

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Answer 1 · 2016-09-01T19:50:50

#= 1/2e^(x^2) ( x^2 - 1 ) + C#

throughout , remember that #d/dx(e^(x^2) )= 2x e^(x^2) #

So:
#int x^3 e^(x^2 ) dx#

setting it up for IBP
#= int color(red)(x^2) d/dx(color(blue)(1/2 e^(x^2 ))) dx#

so by IBP
#= color(red)(x^2) (color(blue)(1/2 e^(x^2 )) ) - int d/dx( color(red)(x^2)) (color(blue)(1/2 e^(x^2 ))) dx#

#= 1/2 x^2 e^(x^2) - int 2 x * 1/2 e^(x^2 ) dx#

#= 1/2 x^2 e^(x^2) - int x e^(x^2 ) dx#

#= 1/2 x^2 e^(x^2) - int 1/2 d/dx( e^(x^2 )) dx#

#= 1/2 x^2 e^(x^2) - 1/2 e^(x^2 ) + C#

#= 1/2e^(x^2) ( x^2 - 1 ) + C#