If $\int \frac{x^{2} (x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x = \frac{- x^{2}}{x \tan x + 1} + f (x) + C$
Then f(x)=

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$2 \log | x \sin x + \cos x | + C$

$\log | x \cos x + \sin x | + C$

$2 \log | x \cos x + \sin x | + C$

$\log | x \sin x + \cos x | + C$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{(x \tan x + 1)^{2}} d x \\ = \int x^{2} \times \frac{(x \sec^{2} x + \tan x)}{(x \tan x + 1)^{2}} d x \\ = \frac{- x^{2}}{(x \tan x + 1)} + \int \frac{2 x}{(x \tan x + 1)} d x \\ I_{1} = \int \frac{2 x}{(x \tan x + 1)} d x \\ = 2 \int \frac{x \cos x}{(x \sin x + \cos x)} d x \end{array}$
Let $x \sin x + \cos x = t$
$x \cos x d x = d t$
$\begin{array}{l} ∴ I_{1} = \int \frac{2 d t}{t} = 2 \log t \\ I_{1} = 2 \log | x \sin x + \cos x | \\ I = \frac{- x^{2}}{x \tan x + 1} + 2 \log | x \sin x + \cos x | + C \\ ∴ f (x) = 2 \log | x \sin x + \cos x | \end{array}$