How do you find the antiderivative of int x^2/(4-x^2) dxx24x2dx?

1 Answer
Jul 29, 2018

int x^2/(4-x^2)dx = -x - ln abs (x-2) + ln abs (x+2) +Cx24x2dx=xln|x2|+ln|x+2|+C

Explanation:

Add and subtract 44 to the numerator:

int x^2/(4-x^2)dx = -int (x^2-4+4)/(x^2-4)dx x24x2dx=x24+4x24dx

int x^2/(4-x^2)dx = -int( 1+4/(x^2-4))dx x24x2dx=(1+4x24)dx

using the linearity of the integral:

int x^2/(4-x^2)dx = -int dx - 4int dx/(x^2-4)x24x2dx=dx4dxx24

int x^2/(4-x^2)dx = -x - 4int dx/(x^2-4)x24x2dx=x4dxx24

Decompose now the resulting integrand using partial fractions:

4/(x^2-4) = 4/((x-2)(x+2))4x24=4(x2)(x+2)

4/(x^2-4) = A/(x-2)+B/(x+2)4x24=Ax2+Bx+2

4/(x^2-4) = (A(x+2)+B(x-2))/((x-2)(x+2))4x24=A(x+2)+B(x2)(x2)(x+2)

4 = Ax +2A +Bx -2B4=Ax+2A+Bx2B

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

{(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