Question

Integrate the function: \(\frac{x+2}{\sqrt{4x-x^2}}\)

Answer 1

Let x+2 = A\(\frac{d}{dx}\)(4x-x²)+B
⇒ x+2 = A(4-2x)+B
Equating the coefficients of x and constant term on both sides, we obtain
-2A = 1 ⇒ A = \(\frac{-1}{2}\)
4A + B = 2 ⇒ B =4
⇒ (x+2) =\(\frac{-1}{2}\)(4-2x)+4
∴ ∫\(\frac{x+2}{\sqrt{4x-x^2}}\) dx = ∫-1/2(4-2x)+4/√4x-x2 dx
= \(\frac{-1}{2}\)∫\(\frac{4-2x}{\sqrt{4x-x^2}}\) dx+4 ∫\(\frac{1}{\sqrt{4x-x^2}}\) dx
Let I₁ = ∫\(\frac{4-2x}{\sqrt{4x-x^2}}\) dx and I₂ =∫\(\frac{1}{\sqrt{4x-x^2}}\) dx
∴ ∫x+2/√4x-x² dx = \(\frac{-1}{2}\) I₁+4I₂                                              ...(1)
then, I₁ = ∫4-2x/√4x-x2 dx
Let 4x-x2 = t
⇒ (4-2x)dx = dt
⇒I₁ = ∫dt/√t = 2√t = 2√4x-x2                                               ...(2)
 I₂ = ∫1/√4x-x2 dx
⇒ 4x-x2 = -(-4x+x2)
=(-4x+x2+4-4)
=4-(x-2)2
=(2)2-(x-2)2

∴ I₂ = ∫1/√(2)2-(x-2)² dx = sin^-1\((\frac{x-2}{2})\)                                   ...(3)

Using equations (2) and (3) in (1), we obtain

∫x+2/√4x-x2 dx = -1/2(2√4x-x2)+4sin-1\((\frac{x-2}{2})\)+C

= \(-\sqrt{4x-x^2}+4sin^{-1}\)\((\frac{x-2}{2})\))+C