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?
 A: 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,
$\int (\frac{1}{x+2}-\frac{1}{x+3})\,dx$
A: Nope! That's not how integration works.
You want to split the denominator using partial fractions.
$\displaystyle\int \frac{1}{(x+3)(x+2)}\,dx =\int \left(\frac{1}{x+2}-\frac{1}{x+3}\right)\,dx=\ln|x+2|-\ln|x+3|+C=\boxed{\ln \left|\frac{x+2}{x+3}\right|+C}$
A: The method you need to use is Partial Fractions.
$\frac{1}{(x+3)(x+2)} =\frac{A}{x+2}+\frac{B}{x+3}$
The LCD is (x+2)(x+3.)
$\frac{1}{(x+3)(x+2)} =\frac{A(x+3) + B(x+2)}{(x+2)(x+3)}$
$$\\$$
Then you set both of the numerators equal to each other:
1 = A(x+3) + B(x+2)
$$\\$$
When does x+3 = 0?
x= -3
Plug -3 into x:
1 = $\require{cancel} \cancel{A(-3+3)}$ + B(-3+2)
$\implies$
1 = -1B
$\implies$
-1 = B
$$\\$$
When does x+2 = 0?
x= -2
Plug -2 into x (and -1 into B because we solved for B above):
1 = A(-2+3) + $\require{cancel} \cancel{-1(-2+2)}$
$\implies$
1 = 1A
$\implies$
1 = A
$$\\$$
Now plug in A= 1 and B= -1 into: $\frac{A}{x+2}+\frac{B}{x+3}$.
Thus your integral is $\int (\frac{1}{x+2}+\frac{-1}{x+3})\,dx$.
When you integrate you get: 
$\boxed{\ln \left|{x+2}| - \ln|x+3\right|+C}$
