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

Question

Evelyn Agard · Accepted Answer

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

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#

Ezra Adams · Answer

undefined To find the antiderivative of ∫ x^2/(4-x^2) dx, you can use partial fraction decomposition. Split the integrand into two fractions with unknown coefficients, A and B, such that:

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

Then, solve for A and B by equating numerators:

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

Solve for A and B:

A = 1/4
B = -1/4

Now integrate each fraction separately:

∫(1/4) * (1/(2+x)) dx + ∫(-1/4) * (1/(2-x)) dx

This results in:

(1/4) * ln|2+x| - (1/4) * ln|2-x| + C

Where C is the constant of integration.