Question

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

Answer 1

$\frac{1}{(x+3)(x+1)}=\frac{1}{2(x+1)}-\frac{1}{2(x+3)}=\int \frac{1}{2(x+1)}-\frac{1}{2(x+3)}dx=\int \frac{1}{2(x+1)}dx-\int \frac{1}{2(x+3)}dx=\frac{1}{2}\cdot \int \frac{1}{x+1}dx-\frac{1}{2}\cdot \int \frac{1}{x+3}dx=\frac{1}{2}\ln |x+1|-\frac{1}{2}\ln |x+3|$

Computing the boundaries:

${\int }_1^2\frac{1}{(x+3)(x+1)}dx=\frac{\ln \left(3\right)-\ln \left(5\right)}{2}-(\frac{\ln \left(2\right)}{2}-\ln \left(2\right))=\frac{\ln \left(3\right)-\ln \left(5\right)}{2}-(\frac{\ln \left(2\right)}{2}-\ln \left(2\right))=\frac{\ln \left(\frac{3}{10}\right)}{2}+\ln \left(2\right)$