int frac{cos x+x sin x}{x(x+cos x)} d x= | vedclass

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(B) Let $I = \int \frac{\cos x + x \sin x}{x(x + \cos x)} dx$.We can rewrite the numerator as: $\cos x + x \sin x = (x + \cos x) + (x \sin x - x) = (x

Vedclass · Accepted Answer

$\log \left|\frac{x}{x+\cos x}\right|+c$

Vedclass · Answer

(B) Let $I = \int \frac{\cos x + x \sin x}{x(x + \cos x)} dx$.We can rewrite the numerator as: $\cos x + x \sin x = (x + \cos x) + (x \sin x - x) = (x + \cos x) - x(1 - \sin x)$.Then,$I = \int \frac{(x + \cos x) - x(1 - \sin x)}{x(x + \cos x)} dx$.Splitting the integral: $I = \int \frac{x + \cos x}{x(x + \cos x)} dx - \int \frac{x(1 - \sin x)}{x(x + \cos x)} dx$.$I = \int \frac{1}{x} dx - \int \frac{1 - \sin x}{x + \cos x} dx$.Let $u = x + \cos x$,then $du = (1 - \sin x) dx$.Substituting these into the integral: $I = \ln |x| - \int \frac{1}{u} du$.$I = \ln |x| - \ln |u| + C$.$I = \ln |x| - \ln |x + \cos x| + C$.$I = \ln \left| \frac{x}{x + \cos x} \right| + C$.

$\int \frac{\cos x+x \sin x}{x(x+\cos x)} d x=$

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