Answer: (a) \(\displaystyle \frac{3(x+2)}{x(x+3)}\); (b) \(\displaystyle \frac13\ln(x^3+3x^2)+C\).

Use the laws of logarithms and exponentials first, then check that any logarithm arguments are positive where required.

(a) Differentiate using

\(\frac{d}{dx}\ln u=\frac{u'}{u}.\)

Here

\(u=x^3+3x^2,\)

so

\(u'=3x^2+6x.\)

Therefore

\(\frac{d}{dx}\ln(x^3+3x^2) =\frac{3x^2+6x}{x^3+3x^2}.\)

Factorising gives

\(\frac{3x(x+2)}{x^2(x+3)} =\frac{3(x+2)}{x(x+3)}.\)

(b) From part (a),

\(\frac{d}{dx}\ln(x^3+3x^2) =\frac{3(x+2)}{x(x+3)}.\)

Hence

\(\frac{x+2}{x(x+3)} =\frac13\frac{d}{dx}\ln(x^3+3x^2).\)

Therefore

\(\int \frac{x+2}{x(x+3)}\,dx =\frac13\ln(x^3+3x^2)+C.\)