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

1 Answer
Jan 3, 2016

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