Question

The value of $\underset{1}{\overset{2}{\int }}\frac{dx}{(x+1)(x+2)}$ is:

• A. $\ln \left(\frac{9}{8}\right)$
• B. $\ln \left(\frac{3}{4}\right)$
• C. $0$
• D. $\ln \left(\frac{3}{5}\right)$

Answer 1

The correct option is A $\ln \left(\frac{9}{8}\right)$

Let $I=\underset{1}{\overset{2}{\int }}\frac{dx}{(x+1)(x+2)}$
Using partial fraction:
$\frac{1}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$
Solving we get $A=1,B=-1$
$\therefore \frac{1}{(x+1)(x+2)}=\frac{1}{(x+1)}-\frac{1}{(x+2)}$
$\Rightarrow I=\underset{1}{\overset{2}{\int }}(\frac{1}{(x+1)}-\frac{1}{(x+2)})dx={\left[\ln \right(x+1)-\ln (x+2\left)\right]}_1^2$
$=\ln 3-\ln 4-\ln 2+\ln 3=\ln \left(\frac{9}{8}\right)$