Q68 / 100 · BITSAT 2023 (Memory Based Paper 1) · Mathematics
$\int \frac{x+3}{(x+4)^2} e^x d x$ is equal to
- Correct answerA$e^x\left(\frac{1}{x+4}\right)+C$
- B$e^{-x}\left(\frac{1}{x+4}\right)+C$
- C$e^{-x}\left(\frac{1}{x-4}\right)+C$
- D$e^{2 x}\left(\frac{1}{x-4}\right)+C$
Solution
Suppose,
$I=\int \frac{x+3}{(x+4)^2} e^x d x=\int \frac{(x+4) 1}{(x+4)^2} e^x d x$
$=\int \frac{e^x}{(x+4)}-\int \frac{e^x}{(x+4)^2} d x$
$=\int e^x\left(\frac{1}{(x+4)}-\frac{1}{(x+4)^2}\right) d x$
$=e^x \frac{1}{(x+4)}+C$
$\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C\right]$
$I=\int \frac{x+3}{(x+4)^2} e^x d x=\int \frac{(x+4) 1}{(x+4)^2} e^x d x$
$=\int \frac{e^x}{(x+4)}-\int \frac{e^x}{(x+4)^2} d x$
$=\int e^x\left(\frac{1}{(x+4)}-\frac{1}{(x+4)^2}\right) d x$
$=e^x \frac{1}{(x+4)}+C$
$\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C\right]$
Hence, the correct answer is option: 'A'.