Q68 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics
The integral $\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x$ is equal to
- A$-\frac{x^2}{x \tan x+1}+c$
- B$2 \log _e|x \sin x+\cos x|+c$
- Correct answerC$-\frac{x^2}{x \tan x+1}+2 \log _e|x \sin x+\cos x|+c$
- D$\frac{x^2}{x^2 \tan x-1}-2 \log _e|x \sin x+\cos x|+c$
Solution
(c) We note that
$\frac{d}{d x}(x \tan x+1)=x \sec ^2 x+\tan x$
$\therefore$ integrating by parts with $x^2$ as first function, we get
$\begin{aligned}
& I=\int x^2 \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x \\
& =x^2\left(-\frac{1}{x \tan x+1}\right)-\int 2 x\left(-\frac{1}{x \tan x+1}\right) d x \\
& =-\frac{x^2}{x \tan x+1}+2 \int \frac{x \cos x}{x \sin x+\cos x} d x \\
& =-\frac{x^2}{x \tan x+1}+2 \log _e|x \sin x+\cos x|+c \\
& \quad\left(\because \frac{d}{d x}(x \sin x+\cos x)=x \cos x\right)
\end{aligned}$
Hence, the correct answer is option: 'C'.