求不定积分 $\int\frac{2(1-u)}{2u+u^2 e^{-u}}du$
求不定积分
\[\int\frac{2(1-u)}{2u+u^2 e^{-u}}du\]
Note that
\[
\frac{2(1-u)}{2u+u^2 e^{-u}}=\frac{1}{u}-\frac{e^{-u}+2}{ue^{-u}+2}
\]
Answer
求不定积分
\[\int\frac{2(1-u)}{2u+u^2 e^{-u}}du\]
Note that
\[
\frac{2(1-u)}{2u+u^2 e^{-u}}=\frac{1}{u}-\frac{e^{-u}+2}{ue^{-u}+2}
\]