How do you find the antiderivative of int x^2/(4-x^2) dx?

1 Answer
Jul 29, 2018

int x^2/(4-x^2)dx = -x - ln abs (x-2) + ln abs (x+2) +C

Explanation:

Add and subtract 4 to the numerator:

int x^2/(4-x^2)dx = -int (x^2-4+4)/(x^2-4)dx

int x^2/(4-x^2)dx = -int( 1+4/(x^2-4))dx

using the linearity of the integral:

int x^2/(4-x^2)dx = -int dx - 4int dx/(x^2-4)

int x^2/(4-x^2)dx = -x - 4int dx/(x^2-4)

Decompose now the resulting integrand using partial fractions:

4/(x^2-4) = 4/((x-2)(x+2))

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

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

4 = Ax +2A +Bx -2B

4 = (A+B)x +2(A-B)

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

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

So:

int x^2/(4-x^2)dx = -x - int dx/(x-2) + int dx/(x+2)

int x^2/(4-x^2)dx = -x - ln abs (x-2) + ln abs (x+2) +C