Question

Evaluate the integral: \[ \int \frac{x^2 (x \sec^2 x + \tan x)}{(x \tan x + 1)^2} dx \]

• A. \( -\frac{x^2}{x \tan x + 1} \)
• B. \( 2 \log_e |x \sin x + \cos x| + C \)
• C. \( -\frac{x^2}{x \tan x + 1} + 2 \log_e |x \sin x + \cos x| + C \)
• D. \( -\frac{x^2}{x \tan x + 1} - 2 \log_e |x \sin x + \cos x| + C \)

Hint:

Use integration by parts along with trigonometric identities to simplify complex integrals.

Answer 1

The Correct Option is C

We note that: \[ \frac{d}{dx} (x \tan x + 1) = x \sec^2 x + \tan x. \] \[\frac{d}{dx}(x \tan x + 1) = x \sec^2 x + \tan x\] integrating by parts with \(x^2\) as the first function, we get \[I = \int x^2 \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx\] \[= x^2 \left(-\frac{1}{x \tan x + 1}\right) - \int 2x \left(-\frac{1}{x \tan x + 1}\right) dx\] \[= -\frac{x^2}{x \tan x + 1} + 2 \int \frac{x}{x \tan x + 1} dx\] \[= -\frac{x^2}{x \tan x + 1} + 2 \int \frac{x \cos x}{x \sin x + \cos x} dx\] \[= -\frac{x^2}{x \tan x + 1} + 2 \log_e |x \sin x + \cos x| + c\] \[\left(\because \frac{d}{dx}(x \sin x + \cos x) = x \cos x \right)\]