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xxln(ex)dx is equal to 

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a
xx+c
b
x.ln x+c
c
(ln x)x+c
d
xln x+c

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detailed solution

Correct option is A

I=∫xxln⁡(ex)dx=∫xx(1+ln⁡x)dx let t=xx=exln⁡x⇒dtdx=xxx⋅1x+ln⁡x=xx1+ln⁡x⇒ dt=xx(1+ln⁡x)dx∴ I=∫dt=t+C=xx+C


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