calculus - Integrate $int frac {1}{(x+2)(x+3)} textrm {dx}$

Integrate $\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$

My Attempt:
$$\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$$
$$\int \dfrac {1}{x+2} \textrm {dx} . \int \dfrac {1}{x+3} \textrm {dx}$$
$$\dfrac {\textrm {log (x+2)}}{1} . \dfrac {\textrm {log (x+3)}}{1} + C$$
$$\textrm {log} (x+2) . \textrm {log} (x+3) + C$$

Is this correct? Or, How do I proceed the other way?

Answer

Method to do -

$\frac{1}{(x+3)(x+2)} =\frac{A}{x+2}+\frac{B}{x+3}$

$1 = A(x+3)+B(x+2)$

Case 1 -

When $x+3=0$

$x=-3$

Put $x=-3$

$1 = -B$

$B = -1$

Case 2 -

When $x+2=0$

$x=-2$

Put $x=-2$

$1 = A$

$A = 1$

Now your integral becomes,

$\int (\frac{1}{x+2}-\frac{1}{x+3})\,dx$