$\int \frac{\sin x - x\cos x}{x(x + \sin x)}dx =$ (where C is an arbitrary constant)

$\int \dfrac{\sin x - x\cos x}{x(x + \sin x)}\,dx$

Our integrand has a product $x(x + \sin x)$ in the denominator. Let's try splitting it into two simpler fractions:

$\dfrac{1}{x} - \dfrac{1 + \cos x}{x + \sin x}$

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Taking the common denominator $x(x + \sin x)$:

$\dfrac{1}{x} - \dfrac{1 + \cos x}{x + \sin x}$

$= \dfrac{(x + \sin x) - x(1 + \cos x)}{x(x + \sin x)}$

$= \dfrac{x + \sin x - x - x\cos x}{x(x + \sin x)}$

$= \dfrac{\sin x - x\cos x}{x(x + \sin x)}$

This matches our integrand, so:

$\int \dfrac{\sin x - x\cos x}{x(x + \sin x)}\,dx = \int \dfrac{1}{x}\,dx - \int \dfrac{1 + \cos x}{x + \sin x}\,dx$

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First term:

$\int \dfrac{1}{x}\,dx = \ln|x|$

Second term — notice that $\dfrac{d}{dx}(x + \sin x) = 1 + \cos x$, so this is of the form $\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)|$:

$\int \dfrac{1 + \cos x}{x + \sin x}\,dx = \ln|x + \sin x|$

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Combining:

$= \ln|x| - \ln|x + \sin x| + C$

Using $\ln a - \ln b = \ln\left(\dfrac{a}{b}\right)$:

$= \ln\left|\dfrac{x}{x + \sin x}\right| + C$