${\int }_0^1\frac{x^2{e}^x}{(x+2{)}^2}dx$

Solution

1. Rewrite the numerator as $$\begin{array}{rl}{x}^2& =(x+2-2{)}^2\\ & =(x+2{)}^2-4(x+2)+4\end{array}$$.
2. Substitute this into the integral to obtain ${\int }_0^1{e}^x\left(1-\frac{4}{x+2}+\frac{4}{(x+2{)}^2}\right)dx$.
3. Recognize that the integrand is in the form $e^x(f(x)+f^{\prime }(x))$ where $f(x)=1-\frac{4}{x+2}$, so the antiderivative is $e^x(1-\frac{4}{x+2})$.
4. Evaluate the definite integral: $$\begin{array}{rl}{\left[e^x\left(1-\frac{4}{x+2}\right)\right]}_0^1& =e(1-\frac{4}{3})-1(1-\frac{4}{2})\\ & =e(-\frac{1}{3})-1(-1)\\ & =\frac{e}{3}-\frac{1}{2}\end{array}$$.
5. Hence the answer is (A).