Question

Evaluate $\int \frac{(x-1)e^x}{(x+1{)}^3}dx$.

• A. $0$
• B. $\frac{(x+1{)}^2}{e^x}+C$
• C. $\frac{e^x}{(x+1{)}^2}+C$
• D. $\frac{e^x}{(x-1{)}^2}+(x+1{)}^2+C$

Answer 1

Answer: (C)

$\frac{e^x}{(x+1{)}^2}+C$

Let $I=\int \frac{(x-1)e^x}{(x+1{)}^3}dx$
$\Rightarrow I=\int \left\{\frac{x+1-2}{(x+1{)}^3}\right\}{e}^{x}dx$
$=\int \{\frac{1}{(x+1{)}^2}-\frac{2}{(x+1{)}^3}\}{e}^{x}dx$
$=\int e^x\cdot \frac{1}{(x+1{)}^2}dx-2\int e^x\frac{1}{(x+1{)}^3}dx$
Applying integrating by parts, we get
$=(\frac{1}{(x+1{)}^2}{e}^x-\int e^x\frac{(-2)}{(x+1{)}^3}dx)-2\int e^x\frac{1}{(x+1{)}^3}dx$
$=\frac{e^x}{(x+1{)}^2}+C$
Note We can use the formula
$\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+C$.