63. determine ∫(xe^x)/((x + 1)^2)dx.

Explanation:

Step1: Rewrite the integrand

Rewrite $xe^x$ as $(x + 1-1)e^x=(x + 1)e^x - e^x$. So the integral $\int\frac{xe^x}{(x + 1)^2}dx=\int\frac{(x + 1)e^x - e^x}{(x + 1)^2}dx=\int\frac{(x + 1)e^x}{(x + 1)^2}dx-\int\frac{e^x}{(x + 1)^2}dx=\int\frac{e^x}{x + 1}dx-\int\frac{e^x}{(x + 1)^2}dx$.

Step2: Use integration - by - parts

Let $u = e^x$ and $dv=\frac{1}{(x + 1)^2}dx$. Then $du=e^x dx$ and $v=-\frac{1}{x + 1}$. By the integration - by - parts formula $\int u dv=uv-\int v du$, we have $\int\frac{e^x}{(x + 1)^2}dx=-\frac{e^x}{x + 1}+\int\frac{e^x}{x + 1}dx$.

Step3: Substitute the result of integration - by - parts

Substitute the result of integration - by - parts into the integral from Step1: $\int\frac{xe^x}{(x + 1)^2}dx=\int\frac{e^x}{x + 1}dx-(-\frac{e^x}{x + 1}+\int\frac{e^x}{x + 1}dx)$. Simplify the right - hand side: $\int\frac{xe^x}{(x + 1)^2}dx=\frac{e^x}{x + 1}+C$.

Answer:

$\frac{e^x}{x + 1}+C$