-\int xe^{(1+i)x}\mbox{d}x=\frac{e^{(i+1)x}\left[1-\left(i+1\right)x\right]}{\left(1+i\right)^{2}}+C= \frac{e^{x}}{2}\left(ie^{ix}(1+i)x-ie^{ix}\right)+C
\int e^{-x}\sin^{2}x\mbox{d}x
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