Let I(x) = int frac{x^2(x sec^2 x + tan x)}{( | vedclass

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(C) We are given $I(x) = \int \frac{x^2(x \sec^2 x + 	an x)}{(x 	an x + 1)^2} dx$. Using integration by parts,let $u = x^2$ and $dv = \frac{x \sec^2

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$\log_e \frac{(\pi+4)^2}{32} - \frac{\pi^2}{4(\pi+4)}$

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(C) We are given $I(x) = \int \frac{x^2(x \sec^2 x + 	an x)}{(x 	an x + 1)^2} dx$. Using integration by parts,let $u = x^2$ and $dv = \frac{x \sec^2 x + 	an x}{(x 	an x + 1)^2} dx$. Note that the derivative of $(x 	an x + 1)$ is $(x \sec^2 x + 	an x)$. Thus,$v = \int (x 	an x + 1)^{-2} d(x 	an x + 1) = -\frac{1}{x 	an x + 1}$. Applying the formula $\int u dv = uv - \int v du$: $I(x) = -\frac{x^2}{x 	an x + 1} + \int \frac{2x}{x 	an x + 1} dx$. $I(x) = -\frac{x^2}{x 	an x + 1} + 2 \int \frac{x \cos x}{x \sin x + \cos x} dx$. Since $\frac{d}{dx}(x \sin x + \cos x) = \sin x + x \cos x - \sin x = x \cos x$,the integral becomes $2 \ln |x \sin x + \cos x| + C$. $I(x) = -\frac{x^2}{x 	an x + 1} + 2 \ln |x \sin x + \cos x| + C$. Given $I(0) = 0$,we have $0 = -0 + 2 \ln |0 + 1| + C \implies C = 0$. At $x = \frac{\pi}{4}$,$I(\frac{\pi}{4}) = -\frac{(\pi/4)^2}{(\pi/4)(1) + 1} + 2 \ln |(\pi/4)(1/\sqrt{2}) + (1/\sqrt{2})| = -\frac{\pi^2/16}{(\pi+4)/4} + 2 \ln |\frac{\pi+4}{4\sqrt{2}}|$. $I(\frac{\pi}{4}) = -\frac{\pi^2}{4(\pi+4)} + \ln |\frac{(\pi+4)^2}{32}| = \log_e \frac{(\pi+4)^2}{32} - \frac{\pi^2}{4(\pi+4)}$.

Let $I(x) = \int \frac{x^2(x \sec^2 x + \tan x)}{(x \tan x + 1)^2} dx$. If $I(0) = 0$,then $I(\frac{\pi}{4})$ is equal to:

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