Question

How do you evaluate `\int \frac { 1} { ( x + 3) ( x + 2) } d x`?

Answer 1

`int \ 1/((x+3)(x+2)) \ dx = -ln|x+3| + ln|x+2| + c`

Explanation:

We seek:

`I = int \ 1/((x+3)(x+2)) \ dx`

We can decompose the integrand into partial fraction, which will take the form:

`1/((x+3)(x+2)) -= A/(x+3) + B/(x+2)`
`" " -= (A(x+2) + B(x+3))/( (x+3)(x+2) )`

Leading to the identity:

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

Where `A,B` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x = -3 => 1 = -A => A = -1`
Put `x = -2 => 1 = B \ \ \ \ \ => B = 1`

So using partial fraction decomposition we have:

`I = int -1/(x+3) + 1/(x+2) \ dx`

And now all integrands are readily integratable, so:

`I = -ln|x+3| + ln|x+2| + c`