If int frac{x}{x tan x+1} , dx = log f(x) + k | vedclass

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(C) Let $I = \int \frac{x}{x 	an x + 1} \, dx = \int \frac{x \cos x}{x \sin x + \cos x} \, dx$. Let $u = x \sin x + \cos x$. Then $du = (\sin x + x \

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$\frac{\pi + 4}{4 \sqrt{2}}$

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(C) Let $I = \int \frac{x}{x 	an x + 1} \, dx = \int \frac{x \cos x}{x \sin x + \cos x} \, dx$. Let $u = x \sin x + \cos x$. Then $du = (\sin x + x \cos x - \sin x) \, dx = x \cos x \, dx$. Thus,$I = \int \frac{1}{u} \, du = \log |u| + k = \log |x \sin x + \cos x| + k$. Comparing this with $\log f(x) + k$,we get $f(x) = |x \sin x + \cos x|$. Now,evaluate $f\left(\frac{\pi}{4}ight) = |\frac{\pi}{4} \sin\left(\frac{\pi}{4}ight) + \cos\left(\frac{\pi}{4}ight)|$. $f\left(\frac{\pi}{4}ight) = |\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}| = |\frac{\pi + 4}{4 \sqrt{2}}| = \frac{\pi + 4}{4 \sqrt{2}}$.

If $\int \frac{x}{x \tan x+1} \, dx = \log f(x) + k$,then $f\left(\frac{\pi}{4}\right) =$

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