Antidifferentiation Question?

∫ 1/(1 - 4x^2) dx
= ∫ 1/[(1 + 2x)(1 - 2x)] dx

1/[(1 + 2x)(1 - 2x)] = A/(1 + 2x) + B/(1 - 2x) (partial fraction)
1 = A(1 - 2x) + B(1 + 2x)
0x + 1 = (2B - 2A)x + (A + B)

0 = 2B - 2A
A = B

1 = A + B
A = B = 1/2

= ∫ 1/[2(1 + 2x)] + 1/[2(1 - 2x)] dx
= 1/4 * [ln|2 + 4x| + ln|2 - 4x|] + C
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∫ x^2/(1 - 4x^2) dx
= ∫ x^2/(1^2 - (2x)^2) dx

Let 2x/1 = sinθ (trigonometric substitution)
dx/dθ = 1/2 * cosθ
dx = 1/2 * cosθ dθ

= ∫ (1/4 * sin^2θ)/(1 - sin^2θ) * 1/2 * cosθ dθ
= ∫ 1/4 * tan^2θ * 1/2 * cosθ dθ
= ∫ 1/8 * sin^2θ/cosθ dθ
= 1/8 * ∫ (1 - cos^2θ)/cosθ dθ
= 1/8 * ∫ secθ - cosθ dθ
= 1/8 * (ln|secθ + tanθ| - sinθ) + C
= 1/8 * (ln|1/√(1 - 4x^2) + 2x/√(1 - 4x^2)| - 2x) + C
= 1/8 * (ln|1 + 2x| - ln|√(1 - 4x^2)| - 2x) + C
= 1/8 * (ln|1 + 2x| - 1/2 * (ln|1 - 4x^2| - 2x)) + C
= 1/8 * (ln|1 + 2x| - 1/2 * (ln|1 + 2x| + ln|1 - 2x| - 2x)) + C
= 1/8 * (1/2 * (ln|1 + 2x| - ln|1 - 2x|) - 2x) + C