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