∫(sin x-xcos x)/(x(x+sin x)) dx = (where C is an arbitrary constant)

The correct answer is Option (3) → \(\log\left|\displaystyle\frac{x}{x+\sin x}\right|+C\)

The numerator can be written in terms of the factors of the denominator:

    \(\sin x - x\cos x = (x + \sin x) - x(1 + \cos x)\)

So the integrand splits into two fractions:

    \(\displaystyle \frac{\sin x - x\cos x}{x(x + \sin x)} = \frac{1}{x} - \frac{1 + \cos x}{x + \sin x}\)

In the second fraction the numerator \(1 + \cos x\) is the derivative of the denominator \(x + \sin x\), so

    \(\displaystyle \int \frac{1 + \cos x}{x + \sin x}\,dx = \log|x + \sin x|\)

Hence

    \(\displaystyle \int \frac{\sin x - x\cos x}{x(x + \sin x)}\,dx = \log|x| - \log|x + \sin x| + C\)

    \(\displaystyle = \log\left|\frac{x}{x + \sin x}\right| + C\)