Q95 / 200 · Manipal (MET) 2012 · Mathematics
Evaluate $\int \frac{(x-1) e^x}{(x+1)^3} d x$
- A0
- B$\frac{(x+1)^2}{e^x}+C$
- Correct answerC$\frac{e^x}{(x+1)^2}+C$
- D$\frac{e^x}{(x-1)^2}+(x+1)^2+C$
Solution
$\begin{aligned}
\text { Let } \quad & \quad I=\int \frac{(x-1) e^x}{(x+1)^3} d x \\
\Rightarrow \quad & \quad I=\int\left\{\frac{x+1-2}{(x+1)^3}\right\} e^x d x \\
& =\int\left\{\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right\} e^x d x \\
& \pm \int e^x \cdot \frac{1}{(x+1)^2} d x-2 \int e^x \frac{1}{(x+1)^3} d x
\end{aligned}$
Applying integrating by parts, we get
$\begin{aligned}
& \begin{array}{l}
=\left(\frac{1}{(x+1)^2} e^x-\int e^x \frac{(-2)}{(x+1)^3} d x\right) -2 \int e^x \frac{1}{(x+1)^3} d x \\
=\frac{e^x}{(x+1)^2}+C
\end{array}
\end{aligned}$
Note We can use the formula
$\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+C$
Hence, the correct answer is option: 'C'.