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

Question

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 1

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,

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

Answer 2

Nope! That's not how integration works.

You want to split the denominator using partial fractions.

$\displaystyle\int \frac{1}{(x+3)(x+2)}\,dx =\int \left(\frac{1}{x+2}-\frac{1}{x+3}\right)\,dx=\ln|x+2|-\ln|x+3|+C=\boxed{\ln \left|\frac{x+2}{x+3}\right|+C}$

Answer 3

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: $\boxed{\ln \left|{x+2}| - \ln|x+3\right|+C}$