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 

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