How do you integrate (9x)/(9x^2+3x-2)9x9x2+3x−2 using partial fractions?

1 Answer
Narad T.
Nov 22, 2016

The answer is ==1/3ln(3x-1)+2/3ln(3x+2)+C==13ln(3x−1)+23ln(3x+2)+C

Explanation:

Let's factorise the denominator

9x^2+3x-2=(3x-1)(3x+2)9x2+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)9x9x2+3x−2=9x(3x−1)(3x+2)=A3x−1+B3x+2

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

Therefore,

9x=A(3x+2)+B(3x-1)9x=A(3x+2)+B(3x−1)

Let x=0x=0, =>⇒, 0=2A-B0=2A−B

Coefficients of xx,

9=3A+3B9=3A+3B, =>⇒, A+B=3A+B=3

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

So,
(9x)/(9x^2+3x-2)=1/(3x-1)+2/(3x+2)9x9x2+3x−2=13x−1+23x+2

int(9xdx)/(9x^2+3x-2)=intdx/(3x-1)+int(2dx)/(3x+2)∫9xdx9x2+3x−2=∫dx3x−1+∫2dx3x+2

=1/3ln(3x-1)+2/3ln(3x+2)+C=13ln(3x−1)+23ln(3x+2)+C

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