Question

How do you integrate `int (x^2)/(x^2 + 3x -4) dx` using partial fractions?

Answer 1

`int x^2/(x^2+3x-4)dx = x+1/5lnabs(x+1) -16/5lnabs(x+4)+C`

Explanation:

Before applying partial fraction decomposition we must reduce the degree of the numerator:

`x^2/(x^2+3x-4) = (x^2+3x-4 -3x + 4)/(x^2+3x-4) = 1- (3x-4)/(x^2+3x-4)`

Factorize the denominator:

`x^2+3x-4 = (x-1)(x+4)`

now decompose in partial fractions:

`(3x-4)/(x^2+3x-4) = A/(x-1)+B/(x+4)`

`(3x-4)/(x^2+3x-4) = (A(x+4) +B(x-1) )/(x^2+3x-4)`

`(3x-4) = (A+B)x + (4A -B)`

`{(A+B=3),(4A-B= -4):}`

`{(A=-1/5),(B= 16/5):}`

`(3x-4)/(x^2+3x-4) = -1/(5(x-1))+16/(5(x+4))`

Now:

`int x^2/(x^2+3x-4)dx = int (1+1/5 1/(x-1) -16/5 1/(x+4))dx`

and using the linearity of the integral:

`int x^2/(x^2+3x-4)dx = int dx+1/5 int dx/(x-1) -16/5 int dx/(x+4)`

`int x^2/(x^2+3x-4)dx = x+1/5lnabs(x+1) -16/5lnabs(x+4)+C`