Question

How to do integrate `int(5x^3-x^2)/(x^2-1)dx` ?

how do you solve by using partial fractions?

Answer 1

`int (5x^3-x^2)/(x^2-1)dx = 5/2 x^2-x+3ln abs(x+1)+2ln abs(x-1)+C`

Explanation:

Reduce the degree of the numerator:

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

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

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

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

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

Use now the partial fraction decomposition to the resulting proper rational function. Factorize the denominator:

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

`(5x-1)/(x^2-1) = A/(x+1)+B/(x-1)`

`(5x-1)/(x^2-1) = (A(x-1)+B(x+1))/(x^2-1)`

`5x-1 = (A+B)x - (A-B)`

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

`{(A= 3),(B=2):}`

Then:

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

Using the linearity of the integral then:

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

`int (5x^3-x^2)/(x^2-1)dx = 5/2 x^2-x+3ln abs(x+1)+2ln abs(x-1)+C`