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?

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