Question

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

Answer 1

$Let\ 2-x = t$

$⇒ -dx = dt$

$⇒$ ∫$$\frac {1}{\sqrt {(2-x)^2+1}}\ dx = $-∫\frac {1}{\sqrt {t^2+1} }dt$

$= -log\ |t+\sqrt {t^2+1}|+C$ $[∫\frac {1}{\sqrt {x^2+a^2} }dt = log\ |x+\sqrt {x^2+a^2}|]$

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

$=log\ |\frac {1}{(2-x)+\sqrt {x^2-4x+5}}|+C$