Question

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

Answer 1

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