Question

Prove ${\int }_1^3\frac{dx}{x^2(x+1)}=\frac{2}{3}+\log \frac{2}{3}$

Answer 1

Let $I={\int }_1^3\frac{dx}{x^2(x+1)}$
Also, let $\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$
$\Rightarrow 1=Ax(x+1)+B(x+1)+C\left(x^2\right)$
$\Rightarrow 1=A{x}^2+Ax+Bx+B+C{x}^2$
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 $B=1$
$\therefore \frac{1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}$
$\Rightarrow I={\int }_1^3\{-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\}dx$
$={[-\log x-\frac{1}{x}+\log (x+1\left)\right]}_1^3$
$={[\log \left(\frac{x+1}{x}\right)-\frac{1}{x}]}_1^3$
$=\log \left(\frac{4}{3}\right)-\frac{1}{3}-\log \left(\frac{2}{1}\right)+1$
$=\log 4-\log 3-\log 2+\frac{2}{3}$
$=\log 2-\log 3+\frac{2}{3}$
$=\log \left(\frac{2}{3}\right)+\frac{2}{3}$
Hence, the given result is proved.