Show that

Let $I=\int_{1}^{3} \frac{d x}{x^{2}(x+1)}$

Also, let $\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$

$\Rightarrow 1=A x(x+1)+B(x+1)+C\left(x^{2}\right)$

$\Rightarrow 1=A x^{2}+A x+B x+B+C x^{2}$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+C=0$

$A+B=0$

$B=1$

On solving these equations, we obtain

$A=-1, C=1$, and $B=1$

$\therefore \frac{1}{x^{2}(x+1)}=\frac{-1}{x}+\frac{1}{x^{2}}+\frac{1}{(x+1)}$

$\Rightarrow I=\int_{1}^{3}\left\{-\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{(x+1)}\right\} d x$

$=\left[-\log x-\frac{1}{x}+\log (x+1)\right]_{1}^{3}$

$=\left[\log \left(\frac{x+1}{x}\right)-\frac{1}{x}\right]_{1}^{3}$

$=\log \left(\frac{4}{3}\right)-\frac{1}{3}-\log \left(\frac{2}{1}\right)+1$

$=\log 4-\log 3-\log 2+\frac{2}{3}$

$=\log 2-\log 3+\frac{2}{3}$

$=\log \left(\frac{2}{3}\right)+\frac{2}{3}$

Hence, the given result is proved.