Using the standard result ∫ex[f(x)+f′(x)]dx=exf(x)+c.
Let f(x)=x1, so f′(x)=−x21.
Then ∫ex(x1−x21)dx=ex⋅x1+c=xex+c.
The value of the integral ∫ex(x1−x21)dx is :
Held on 7 Jun 2023 · Verified 13 Jul 2026.
xex+c
xe−x+c
ex+C
e−x+C
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