How do you integrate #(9x)/(9x^2+3x-2)# using partial fractions?

1 Answer
Nov 22, 2016

The answer is #==1/3ln(3x-1)+2/3ln(3x+2)+C#

Explanation:

Let's factorise the denominator

#9x^2+3x-2=(3x-1)(3x+2)#

Now, we can start by the decomposition into partial fractions

#(9x)/(9x^2+3x-2)=(9x)/((3x-1)(3x+2))=A/(3x-1)+B/(3x+2)#

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

Therefore,

#9x=A(3x+2)+B(3x-1)#

Let #x=0#, #=>#, #0=2A-B#

Coefficients of #x#,

#9=3A+3B#, #=>#, #A+B=3#

Solving for A and B, we get #A=1# and #B=2#

So,
#(9x)/(9x^2+3x-2)=1/(3x-1)+2/(3x+2)#

#int(9xdx)/(9x^2+3x-2)=intdx/(3x-1)+int(2dx)/(3x+2)#

#=1/3ln(3x-1)+2/3ln(3x+2)+C#