Q104 / 180 · KCET 2022 · Mathematics
$\int_0^1 \frac{x e^x}{(2+x)^3} d x$ is equal to
- A$\frac{1}{27} \cdot e-\frac{1}{8}$
- B$\frac{1}{27} \cdot e+\frac{1}{8}$
- C$\frac{1}{9} \cdot e+\frac{1}{4}$
- Correct answerD$\frac{1}{9} \cdot e-\frac{1}{4}$
Solution
Let
Here, $\frac{1}{(2+x)^2}=f(x)$ and $\frac{-2}{(2+x)^3}=f^{\prime}(x)$
We know, $\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c$
$$
l=\left[e^x \cdot \frac{1}{(2+x)^2}\right]_0^1=\frac{e^x}{(2+1)^2}-\frac{e^0}{(2+0)^2}=\frac{e}{9}-\frac{1}{4}
$$
Hence, the correct answer is option: 'D'.