∫xex(1+x)2isequalto:
Find the integral of the given expression
Given: I=∫xex(1+x)2dx
I=∫(x+1-1)ex(1+x)2dx
I=∫ex(1+x)dx-∫ex(1+x)2dx
Now using the integration by parts method in the first integral, we get;
I=ex(1+x)+∫ex(1+x)2dx-∫(ex(1+x)2)dx+CI=ex(1+x)+∫(1(1+x)2)exdx-∫(ex(1+x)2)dx+CI=ex(1+x)+C
Hence, the answer is ex(1+x)+C, where C is an arbitrary constant.