Question

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

Answer 1

${\int }_1^2\frac{x}{(x+1)(x+2)}dxlet,\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\Rightarrow x=(A+B)x+(2A+B)comparingbothsidewegetA+B=1and2A+B=0solvingbothaboveequationwegetA=-1andB=2\therefore {\int }_1^2\frac{x}{(x+1)(x+2)}dx={\int }_1^2[\frac{-1}{x+1}+\frac{2}{x+2}]dx={\left[2\log \right(x+2)-\log (x+1\left)\right]}_1^2={[\log {(x+2)}^2-\log (x+1\left)\right]}_1^2={\left[\log \frac{{(x+2)}^2}{x+1}\right]}_1^2=\log \frac{32}{27}$