Question

How do you use partial fraction decomposition to decompose the fraction to integrate `11/(3x^2-7x-6)`?

Answer 1

`int 11/(3x^2-7x-6)dx = ln abs((x-3)/(3x+2))+C`

Explanation:

Factorize the denominator by solving the equation:

`3x^2-7x-6 = 0`

`x = (7+-sqrt(49+72))/6 = (7+-sqrt(121))/6 = (7+-11)/6={(-2/3),(3):}`

Then:

`3x^2-7x-6 = 3(x+2/3)(x-3) = (3x+2)(x-3)`

Apply partial fractions decomposition:

`11/(3x^2-7x-6) = A/(3x+2)+B/(x-3)`

`11/(3x^2-7x-6) = (A(x-3)+B(3x+2))/(3x^2-7x-6)`

`11 = (A+3B)x - 3A +2B`

`{(A+3B = 0),(-3A+2B=11):}`

`{(A=-3),(B=1):}`

So:

`11/(3x^2-7x-6) = -3/(3x+2)+1/(x-3)`

`int 11/(3x^2-7x-6)dx = -int (3dx)/(3x+2)+int dx/(x-3)`

`int 11/(3x^2-7x-6)dx = -int (d(3x+2))/(3x+2)+int (d(x-3))/(x-3)`

`int 11/(3x^2-7x-6)dx = -ln abs(3x+2)+ln abs(x-3)+C`

`int 11/(3x^2-7x-6)dx = ln abs((x-3)/(3x+2))+C`