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Question

Evaluate xex(x+1)2dx

A
ex[1x+1]+c
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B
ex[1x+1]+c
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C
exx+c
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D
ex[1(x+1)2]+c
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Solution

The correct option is A ex[1x+1]+c
I=xex(x+1)2dx
=(x+11)ex(x+1)2dx
=(x+1)ex(x+1)2dxex(x+1)2dx
=exx+1dxex(x+1)2dx

We know, u.v dx=uv dx (dudxv dx)dx

Now by integrating the functions by parts, for the first part taking ex as u and 1x+1 as v

I=exx+1+ex(x+1)2dxex(x+1)2dx
I=exx+1+C

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