Question

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

Answer 1

Let  ${\displaystyle I=\int \frac{x-1}{\sqrt{x^2-1}} dx}$

${\displaystyle =\int \frac{x}{\sqrt{x^2-1}} dx-\int \frac{1}{\sqrt{x^2-1}} dx}$

${\displaystyle =I_1-I_2}$  (say)

Now , ${\displaystyle I_1=\frac{x}{\sqrt{x^2-1}} dx}$

Put x² - 1 = t

2x dx = dt ⇒ x dx = ${\displaystyle \frac{1}{2}}$ dt

${\displaystyle \therefore I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\frac{1}{2}\int t^{\frac{-1}{2}}dt}$

${\displaystyle =\frac{1}{2}\times \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C_1=\sqrt{t}=C_1}$

${\displaystyle =\sqrt{x^2-1}+C_1}$

and ${\displaystyle I_2\int \frac{1}{\sqrt{x^2-1}} dx}$

${\displaystyle =\log [x+\sqrt{x^2}-1]+C_2}$            ${\displaystyle ....[\because \int \frac{dx}{\sqrt{x^2-a^2}}=\log |x+\sqrt{x^2-a^2}|+C]}$

${\displaystyle \therefore I=\sqrt{x^2-1}-\log |x+\sqrt{x^2-1}|+C}$