$\frac{1 + t a n^{2} x}{1 - t a n^{2} x} d x$ is equal to

l o g \frac{1 - t a n x}{1 + t a n x} + c

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l o g \frac{1 + t a n x}{1 - t a n x} + c

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\frac{1}{2} l o g \frac{1 - t a n x}{1 + t a n x} + c

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\frac{1}{2} l o g \frac{1 + t a n x}{1 - t a n x} + c

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Solution

The correct option is D $\frac{1}{2} l o g \frac{1 + t a n x}{1 - t a n x} + c$
$\int \frac{1 + {tan}^{2} (x)}{1 - {tan}^{2} (x)} d x$ $= \int \frac{{sec}^{2} (x)}{1 - {tan}^{2} (x)} d x$
$A p p l y u - s u b s t i t u t i o n : u = tan (x)$
$\int \frac{1}{1 - u^{2}} d u$
$U s e t h e c o m m o n i n t e g r a l : \int \frac{1}{1 - u^{2}} d u = \frac{ln | u + 1 |}{2} - \frac{ln | u - 1 |}{2}$
$= \frac{ln | u + 1 |}{2} - \frac{ln | u - 1 |}{2}$
$S u b s t i t u t e b a c k u = tan (x)$
$= \frac{ln | tan (x) + 1 |}{2} - \frac{ln | tan (x) - 1 |}{2}$
$A d d a c o n s t a n t t o t h e s o l u t i o n$
$= \frac{ln | tan (x) + 1 |}{2} - \frac{ln | tan (x) - 1 |}{2} + C$
$= \frac{1}{2} l o g \frac{1 + t a n x}{1 - t a n x} + C$

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