Question

Integrate the following functions.
$\int \frac{x-1}{\sqrt{x^2-1}}dx.$

Answer 1

Let $I=\int \frac{x-1}{\sqrt{x^2-1}}dx=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx=I_1-I_2.....\left(i\right)$
Now, $I_1=\int \frac{x}{\sqrt{x^2-1}}dx.$

Let $x^2-1=t\Rightarrow 2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{2x}$
$\therefore I_1=\int \frac{x}{\sqrt{t}}\frac{dt}{2x}=\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\frac{1}{2}\int t^{-\frac{1}{2}}dt=\frac{1}{2}\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right]+C_1=\sqrt{t}+C_1=\sqrt{x^2-1}+C_1(\because t=x^2-1)$
Now, $I_2=\int \frac{1}{\sqrt{x^2-1}}dx$
$=log|x+\sqrt{x^2-1}|+C_2[\because \int \frac{dx}{\sqrt{x^2-a^2}}=log|x+\sqrt{x^2-a^2}|]$
On putting the values of $I_1$ and $I_2$ in Eq. (i), we get
$\int \frac{x-1}{\sqrt{x^2-1}}dx=\sqrt{x^2-1}=log|x+\sqrt{x^2-1}|+C$ (where, $C=C_1-C_2)$