int {frac{{1 + {{tan }^2}x}}{{1 - {{tan }^2}x | vedclass

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(D) Let $I = \int {\frac{{1 + {{	an }^2}x}}{{1 - {{	an }^2}x}}dx} $.Using the identity $1 + 	an^2 x = \sec^2 x$,we get $I = \int {\frac{{{{\sec }^2

Vedclass · Accepted Answer

$\frac{1}{2}\log \left( {\frac{{1 + 	an x}}{{1 - 	an x}}} ight) + c$

Vedclass · Answer

(D) Let $I = \int {\frac{{1 + {{	an }^2}x}}{{1 - {{	an }^2}x}}dx} $.Using the identity $1 + 	an^2 x = \sec^2 x$,we get $I = \int {\frac{{{{\sec }^2}x}}{{1 - {{	an }^2}x}}dx} $.Substitute $	an x = t$,then $\sec^2 x \, dx = dt$.The integral becomes $I = \int {\frac{{dt}}{{1 - {t^2}}}} $.Using the standard formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} ight| + c$,we have:$I = \frac{1}{2(1)} \log \left| \frac{1+t}{1-t} ight| + c$.Substituting $t = 	an x$ back,we get $I = \frac{1}{2} \log \left| \frac{1 + 	an x}{1 - 	an x} ight| + c$.

$\int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}\,dx} $ is equal to

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