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

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Question

(B) Let $I = \int \frac{\cos x + x \sin x}{x(x + \cos x)} dx$.Observe the denominator $f(x) = x^2 + x \cos x$.The derivative of the denominator is $f'

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{\cos x + x \sin x}{x(x + \cos x)} dx$.Observe the denominator $f(x) = x^2 + x \cos x$.The derivative of the denominator is $f'(x) = \frac{d}{dx}(x^2 + x \cos x) = 2x + \cos x - x \sin x$. This does not match the numerator directly.Let us rewrite the integral as:$I = \int \frac{x + \cos x + x \sin x - x}{x(x + \cos x)} dx$$I = \int \left( \frac{x + \cos x}{x(x + \cos x)} + \frac{x \sin x - x}{x(x + \cos x)} \right) dx$$I = \int \left( \frac{1}{x} + \frac{x(\sin x - 1)}{x(x + \cos x)} \right) dx$$I = \int \frac{1}{x} dx + \int \frac{\sin x - 1}{x + \cos x} dx$Let $u = x + \cos x$,then $du = (1 - \sin x) dx$,which means $-(1 - \sin x) dx = du$,or $(\sin x - 1) dx = -du$.$I = \ln|x| - \int \frac{1}{u} du$$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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