Question

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

Answer 1

Let $x+2=A\frac{d}{dx}(x^2-1)+B$ ......(1)
$\Rightarrow x+2=A\left(2x\right)+B$
Equating the coefficients of $x$ and constant term on both sides, we obtain
$2A=1\Rightarrow A=\frac{1}{2}$
$B=2$
From (1), we obtain
$(x+2)=\frac{1}{2}\left(2x\right)+2$
Then, $\int \frac{x+2}{\sqrt{x^2-1}}dx=\int \frac{\frac{1}{2}\left(2x\right)+2}{\sqrt{x^2-1}}dx$
$=\frac{1}{2}\int \frac{2x}{\sqrt{x^2-1}}dx+\int \frac{2}{\sqrt{x^2-1}}dx$ ......(2)
For $\frac{1}{2}\int \frac{2x}{\sqrt{x^2-1}}dx$, let $x^2-1=t\Rightarrow 2xdx=dt$
$\Rightarrow \frac{1}{2}\int \frac{2x}{\sqrt{x^2-1}}dx=\frac{1}{2}\int \frac{dt}{\sqrt{t}}$
$=\frac{1}{2}\left[2\sqrt{t}\right]=\sqrt{t}=\sqrt{x^2-1}$
And, $\int \frac{2}{\sqrt{x^2-1}}dx=2\int \frac{1}{\sqrt{x^2-1}}dx=2\log |x+\sqrt{x^2-1}|$
From equation (2), we obtain
$\int \frac{x+2}{\sqrt{x^2-1}}dx=\sqrt{x^2-1}+2\log |x+\sqrt{x^2-1}|+C$