Question

How do you use partial fraction decomposition to decompose the fraction to integrate `x/((x+7)(x+8)(x+9))`?

Answer 1

`intx/((x+7)(x+8)(x+9))dx=7/2ln|x+7|+8ln|x+8|-9/2ln|x+9|+C`

Explanation:

We can decompose the integrand as follows:

`x/((x+7)(x+8)(x+9))=A/(x+7)+B/(x+8)+C/(x+9)`

Add up the right side:

`x/((x+7)(x+8)(x+9))=(A(x+8)(x+9))/((x+7)(x+8)(x+9))+((B)(x+7)(x+9))/((x+7)(x+8)(x+9))+((C)(x+7)(x+8))/((x+7)(x+8)(x+9))`

Set numerators equal:

`x=A(x+8)(x+9)+B(x+7)(x+9)+C(x+7)(x+8)`

We need to find `A,B,C.` Fortunately, it looks like we can do this by choosing particular values of `x` which will eliminate some terms on the right side.

`x=-8:`

`-8=B(-1)(1), B=8`

`x=-7:`

`-7=A(-1)(2), A=7/2`

`x=-9:`

`-9=C(-2)(-1), C=-9/2`

So, with the decomposed fraction, the integral becomes

`7/2intdx/(x+7)+8intdx/(x+8)-9/2intdx/(x+9)`

Note that all constants were factored out.

These are all simple integrals.

In general, `intdx/(x+-a)=ln|x+-a|+C`

Integrating yields

`intx/((x+7)(x+8)(x+9))dx=7/2ln|x+7|+8ln|x+8|-9/2ln|x+9|+C`