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If $\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x=A \log$ $(|x \sin x+\cos x|)+B \frac{f(x)}{(x \tan x+1)}+C$, then $f(A+B)=$
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Let $I=\int x^2 \cdot \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x$
$=x^2\left(-\frac{1}{(x \tan x+1)}\right)-\int 2 x\left(-\frac{1}{(x \tan x+1}\right) d x$
$\left[\right.$ since $\left.\int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2}=\int \frac{d t}{t^2}=-\frac{1}{t}=-\frac{1}{(x \tan x+1)}\right]$
$=-\left(\frac{x^2}{x \tan x+1}\right)+\int \frac{2 x \cos x}{(x \sin x+\cos x)} d x$
[putting $x \sin x+\cos x=u$ $\Rightarrow(x \cos x+\sin x-\sin x) d x=d u]$
$=\frac{-x^2}{(x \tan x+1)}+2 \int \frac{d u}{u}$
$=\frac{-x^2}{x \tan x+1}+2 \log |u|+C$
$=\frac{-x^2}{x \tan x+1}+2 \log |x \sin x+\cos x|+C$
$\therefore A=2$ and $B=-1$ and $f(x)=x^2$
$\therefore f(A+B)=f(2-1)=f(1)=(1)^2=1$
$=x^2\left(-\frac{1}{(x \tan x+1)}\right)-\int 2 x\left(-\frac{1}{(x \tan x+1}\right) d x$
$\left[\right.$ since $\left.\int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2}=\int \frac{d t}{t^2}=-\frac{1}{t}=-\frac{1}{(x \tan x+1)}\right]$
$=-\left(\frac{x^2}{x \tan x+1}\right)+\int \frac{2 x \cos x}{(x \sin x+\cos x)} d x$
[putting $x \sin x+\cos x=u$ $\Rightarrow(x \cos x+\sin x-\sin x) d x=d u]$
$=\frac{-x^2}{(x \tan x+1)}+2 \int \frac{d u}{u}$
$=\frac{-x^2}{x \tan x+1}+2 \log |u|+C$
$=\frac{-x^2}{x \tan x+1}+2 \log |x \sin x+\cos x|+C$
$\therefore A=2$ and $B=-1$ and $f(x)=x^2$
$\therefore f(A+B)=f(2-1)=f(1)=(1)^2=1$
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