Question

Integrate the function: $\frac {x+2}{\sqrt {x^2-1}}$

Answer 1

Let x+2 = A$\frac {d}{dx}$(x2-1) + B .....(1)

⇒ x+2 = A(2x)+B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 1 ⇒ A = $\frac 12$

B = 2

From (1), we obtain

(x+2) = $\frac 12$(2x)+2

Then, ∫$$\frac {x+2}{\sqrt {x^2-1}}\ dx = $∫\frac {\frac 12(2x)+2}{\sqrt {x^2-1}}$ .....(2)

In $\frac 12 ∫\frac {2x}{\sqrt {x^2-1}}\ dx$ dx, let x2-1 = t ⇒ 2x dx = dt

$\frac 12 ∫\frac {2x}{\sqrt {x^2-1}}\ dx$= $\frac 12 ∫\frac {dt}{\sqrt t}$

= $\frac 12[2\sqrt t]$

=$\sqrt t$

=$\sqrt {x^2-1}$

Then, $∫\frac {2}{\sqrt {x^2-1}}\ dx$ = $2∫\frac {x}{\sqrt {x^2-1}}\ dx$ = $2\ log\ |x+\sqrt {x^2-1}|$

From equation (2), we obtain

∫$$\frac {x+2}{\sqrt {x^2-1}}\ dx = $\sqrt {x^2-1}+2\ log\ |x+\sqrt {x^2-1}|+C$