OpenStudy (anonymous):

I = int (x^3+3x) * e^x dx (using repeated integration by parts)

5 years ago
OpenStudy (ash2326):

\[\int\limits_{}^{}(x^3+3x) e^x dx\] \[=(x^3+3x)e^x-\int\limits_{}^{}(3x^2+3) e^x dx\] \[(x^3+3x)-(3x^2+3)+\int\limits_{}^{}(6x) e^x dx\]

5 years ago
OpenStudy (ash2326):

\[((x^3+3x)-(3x^2+3))e^x+6x e^x-6e^x\]

5 years ago
OpenStudy (ash2326):

we'll get \[(x^3+3x-3x^2-3+6x-x)e^x + C\] \[(x^3-3x^2+8x-3)e^x +c\]

5 years ago
OpenStudy (anonymous):

yes, ash, thats the answer but I don't understand why is is a "+" sign in front of 6x e^x instead of a '-'

5 years ago
OpenStudy (ash2326):

integrate 6x e^x , you'll ge(t 6x-6)e^x

5 years ago
OpenStudy (anonymous):

when we do an integration by parts for the third time, we achieve: |dw:1325937540524:dw|

5 years ago