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

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(D) Let $I = \int \frac{1 + 	an^{2} x}{1 - 	an^{2} x} dx$. Since $1 + 	an^{2} x = \sec^{2} x$,we have $I = \int \frac{\sec^{2} x}{1 - 	an^{2} x} d

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

Vedclass · Answer

(D) Let $I = \int \frac{1 + 	an^{2} x}{1 - 	an^{2} x} dx$. Since $1 + 	an^{2} x = \sec^{2} x$,we have $I = \int \frac{\sec^{2} x}{1 - 	an^{2} x} dx$. Substitute $	an x = t$,then $\sec^{2} x \ dx = dt$. Substituting these into the integral,we get $I = \int \frac{dt}{1 - t^{2}}$. Using the standard integral formula $\int \frac{dx}{a^{2} - x^{2}} = \frac{1}{2a} \log \left| \frac{a + x}{a - x} ight| + c$,we get $I = \frac{1}{2} \log \left| \frac{1 + t}{1 - t} ight| + c$. Replacing $t$ with $	an x$,we obtain $I = \frac{1}{2} \log \left| \frac{1 + 	an x}{1 - 	an x} ight| + c$.

$\int \frac{1 + \tan^{2} x}{1 - \tan^{2} x} dx =$

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