Question

Integrate:

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

Answer 1

Consider the given integral.

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

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

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

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

$I=-[\ln 3-0]-(-\frac{2}{3})+[\ln 4-\ln 2]$

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

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

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

Hence, this is the answer.