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?

Nope! That's not how integration works.
You want to split the denominator using partial fractions.
$\displaystyle\int \frac{1}{(x+3)(x+2)}\,dx =\int \left(\frac{1}{x+2}-\frac{1}{x+3}\right)\,dx=\ln|x+2|-\ln|x+3|+C=\boxed{\ln \left|\frac{x+2}{x+3}\right|+C}$
