$\int x e^{x}\,dx$ equals
A$x e^{x}+e^{x}+C$
B$\dfrac{x^2}{2}e^{x}+C$
C$x e^{x}-e^{x}+C$
D$e^{x}(x-1)^2+C$
Answer & Solution
Correct answer: C. $x e^{x}-e^{x}+C$
1. Take $u=x$ (first), $dv=e^{x}\,dx$, so $v=e^{x}$.
2. By parts: $x e^{x}-\int 1\cdot e^{x}\,dx$.
3. $=x e^{x}-e^{x}+C$.
4. Check: $\dfrac{d}{dx}(xe^{x}-e^{x})=e^{x}+xe^{x}-e^{x}=xe^{x}$, confirming C.
_Source: syllabus Mathematics Ch 7 "Integrals", p.325_
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