Question

Integrate:
${\int }_1^3\frac{dx}{x^2\left(x+1\right)}$

Answer 1

Consider the given integral.

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

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

$I={[-\ln \left(x\right)-\frac{1}{x}+\ln (x+1)]}_1^3$

$I=[(-\ln \left(3\right)-\frac{1}{3}+\ln (3+1))-(-\ln \left(1\right)-\frac{1}{1}+\ln (1+1))]$

$I=-\ln 3-\frac{1}{3}+\ln 4+1-\ln 2$

$I=-\ln 3+\frac{2}{3}+2\ln 2-\ln 2$

$I=-\ln 3+\frac{2}{3}+\ln 2$

$I=\frac{2}{3}+\ln 2-\ln 3$

$I=\frac{2}{3}+\ln \frac{2}{3}$

Hence, this is the answer.