Prove that $\int_{1}^{3} \frac{dx}{x^{2}(x+1)} = \frac{2}{3} + \log \frac{2}{3}$.

Let $I = \int_{1}^{3} \frac{dx}{x^{2}(x+1)}$.
Using partial fractions,let $\frac{1}{x^{2}(x+1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x+1}$.
Multiplying by $x^{2}(x+1)$,we get $1 = Ax(x+1) + B(x+1) + Cx^{2}$.
Expanding the terms,$1 = (A+C)x^{2} + (A+B)x + B$.
Equating coefficients,we get $B = 1$,$A+B = 0 \Rightarrow A = -1$,and $A+C = 0 \Rightarrow C = 1$.
Thus,$\frac{1}{x^{2}(x+1)} = -\frac{1}{x} + \frac{1}{x^{2}} + \frac{1}{x+1}$.
Integrating term by term:
$I = \int_{1}^{3} \left( -\frac{1}{x} + \frac{1}{x^{2}} + \frac{1}{x+1} \right) dx$
$I = \left[ -\log|x| - \frac{1}{x} + \log|x+1| \right]_{1}^{3}$
$I = \left[ \log\left| \frac{x+1}{x} \right| - \frac{1}{x} \right]_{1}^{3}$
$I = \left( \log\left( \frac{4}{3} \right) - \frac{1}{3} \right) - \left( \log(2) - 1 \right)$
$I = \log\left( \frac{4}{3} \right) - \log(2) - \frac{1}{3} + 1$
$I = \log\left( \frac{4/3}{2} \right) + \frac{2}{3}$
$I = \log\left( \frac{2}{3} \right) + \frac{2}{3}$.
Hence,the result is proved.