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#