$\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$ is equal to

A. $\tan x+\cot x+C$

B. $\tan x+\operatorname{cosec} x+C$

C. $-\tan x+\cot x+C$

D. $\tan x+\sec x+C$


$\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x=\int\left(\frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\right) d x$

$=\int\left(\sec ^{2} x-\operatorname{cosec}^{2} x\right) d x$

$=\tan x+\cot x+\mathrm{C}$

Hence, the correct answer is A.

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