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

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(B) Let $I = \int \frac{\cos x + x \sin x}{x(x - \cos x)} dx$. We can rewrite the denominator as $x^2(1 - \frac{\cos x}{x})$. So,$I = \int \frac{\cos

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$\log \left| 1 - \frac{\cos x}{x} \right| + c$

Vedclass · Answer

(B) Let $I = \int \frac{\cos x + x \sin x}{x(x - \cos x)} dx$. We can rewrite the denominator as $x^2(1 - \frac{\cos x}{x})$. So,$I = \int \frac{\cos x + x \sin x}{x^2(1 - \frac{\cos x}{x})} dx$. Let $t = 1 - \frac{\cos x}{x}$. Then,differentiating with respect to $x$ using the quotient rule: $dt = - \left( \frac{x(-\sin x) - \cos x(1)}{x^2} \right) dx = \frac{x \sin x + \cos x}{x^2} dx$. Substituting these into the integral: $I = \int \frac{1}{t} dt = \ln |t| + c$. Substituting back $t = 1 - \frac{\cos x}{x}$: $I = \ln \left| 1 - \frac{\cos x}{x} \right| + c$.

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

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