Question

How do you integrate `int (5x - 4 ) / (x^2 -4x) dx` using partial fractions?

Answer 1

Use partial fractions to give the expression in a form that you can integrate, and then integrate it normally.
Answer: `lnx + 4ln(x-4) + c`

Explanation:

Partial Fraction
Initially, ignore the fact that you will be integrating.
Let
`f(x)=(5x-4)/(x^2-4x)`

Factorise the denominator:

`f(x)=(5x-4)/(x(x-4))`

Then let

`f(x)=A/x + B/(x-4)`

So

`(5x-4)/(x(x-4))=A/x + B/(x-4)`

Then multiply by the denominator of `f(x)`

`5x-4=(x(x-4)A)/x + (x(x-4)B)/(x-4)`

This causes the cancellation of some terms:

`5x-4=A(x-4)+Bx`

To find `A` and `B`, you can use substitutions or compare coefficients. I prefer comparing coefficients for small problems, but using a mixture of the two techniques is often helpful.

Substitution
let `x=4`
Therefore:
`20-4=0+4B`
`16=4B`
`B=4`

Comparing Coefficients
For the constant terms:
`-4=-4A`
`A=1`

Now we have everything we need to form our partial fraction:

`f(x)=A/x + B/(x-4)`

`f(x)=1/x + 4/(x-4)`

We are now ready to integrate.

Integration

`int(5x-4)/(x^2-4x)dx=int1/x + 4/(x-4)dx`

It is easier to see if we separate the two parts of the integration:

`int1/xdx+int4/(x-4)dx`

These both integrate in the same manner, using the standard antiderivative rule:

`d/(dx)(lnx)=1/(x)`

So we get the answer:
`lnx + 4ln(x-4) + c`