int left( frac{4 tan^4 x + 3 tan^2 x - 1}{tan | vedclass

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Question

(C) Let $I = \int \frac{4 	an^4 x + 3 	an^2 x - 1}{	an^2 x + 4} dx$. First, factor the numerator: $4 	an^4 x + 3 	an^2 x - 1 = (4 	an^2 x - 1)(\

Vedclass · Accepted Answer

$4 	an x - \frac{17}{2} 	an^{-1} \left( \frac{	an x}{2} ight) + c$

Vedclass · Answer

(C) Let $I = \int \frac{4 	an^4 x + 3 	an^2 x - 1}{	an^2 x + 4} dx$. First, factor the numerator: $4 	an^4 x + 3 	an^2 x - 1 = (4 	an^2 x - 1)(	an^2 x + 1)$. Since $1 + 	an^2 x = \sec^2 x$, we have $I = \int \frac{(4 	an^2 x - 1) \sec^2 x}{	an^2 x + 4} dx$. Let $u = 	an x$, then $du = \sec^2 x dx$. The integral becomes $I = \int \frac{4u^2 - 1}{u^2 + 4} du$. Perform polynomial division or rewrite the numerator: $I = \int \frac{4(u^2 + 4) - 17}{u^2 + 4} du$. $I = \int \left( 4 - \frac{17}{u^2 + 2^2} ight) du$. Integrating, we get $I = 4u - 17 \cdot \frac{1}{2} 	an^{-1} \left( \frac{u}{2} ight) + C$. Substituting $u = 	an x$ back, $I = 4 	an x - \frac{17}{2} 	an^{-1} \left( \frac{	an x}{2} ight) + C$.

$\int \left( \frac{4 \tan^4 x + 3 \tan^2 x - 1}{\tan^2 x + 4} \right) dx =$

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