Question

Find the integral of $\int \frac{1}{(x+2)(x+3)}dx$

• A. ${\log }_e\frac{(x+2)}{(x+3)}+C$
• B. ${\log }_e(x+2)+{\log }_e(x+3)+C$
• C. ${\log }_e\frac{(x+3)}{(x+2)}+C$
• D. None of these

Answer 1

The correct option is A ${\log }_e\frac{(x+2)}{(x+3)}+C$

Let
$\frac{1}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}$
$1=A(x+3)+B(x+2)\Rightarrow Ax+Bx+3A+2B=1$
Comparing the coefficient of $x$ and constant on both sides,
$A+B=0$ and $3A+2B=1$
Solving the above two equations, we get
$A=1$ and $B=-1$
$\Rightarrow \int \frac{dx}{(x+2)(x+3)}=\int (\frac{1}{(x+2)}-\frac{1}{(x+3)})dx$
$={\log }_e(x+2)–{\log }_e(x+3)+C={\log }_e\frac{x+2}{x+3}+C$