$The integral \int {(\frac{x}{xsinx + cosx})}^{2} dx is equal to$ (where C is a constant of integration) :

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$\sec x + \frac{x \tan x}{x \sin x + \cos x} + C$

$\tan x - \frac{x \sec x}{x \sin x + \cos x} + C$

$\sec x - \frac{x \tan x}{x \sin x + \cos x} + C$

$\tan x + \frac{x \sec x}{x \sin x + \cos x} + C$

answer is A.

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Detailed Solution

$\int \frac{x^{2}}{(x \sin x + \cos x)^{2}} d x M u l t i p l y a n d d i v i d e b y \cos x = \int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x + \cos x)^{2}} d x$

$\begin{array}{l} I n t e g r a t i o n b y p a r t s \\ = \frac{x}{\cos x} \int \frac{x \cos x}{{(x \sin x + \cos x)}^{2}} d x - \int [\frac{d}{d x} \frac{x}{\cos x} - \int \frac{x \cos x}{{(x \sin x + \cos x)}^{2}} d x] d x \\ p u t x \sin x + \cos x = t \\ \Rightarrow (x \cos x + \sin x - \sin x) d x = d t \\ \Rightarrow x \cos x d x = d t \\ = \frac{x}{\cos x} \cdot \frac{- 1}{(x \sin x + \cos x)} + \int \frac{\cos x + x \sin x}{\cos^{2} x} \cdot \frac{1}{(x \sin x + \cos x)} d x \\ = \frac{- x}{\cos x (x \sin x + \cos x)} + \tan x + C \\ = \tan x - \frac{x s e c x}{x \sin x + \cos x} + C \end{array}$