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

Answer

Clearly, we can write, x + 2 = $\frac{1}{2}(2 x)+2$
$\Rightarrow \int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$ 
$= \frac{1}{2} \int \frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}+2 \int \frac{1}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{d} \mathrm{x}$ 
Now in, $I_1$= $\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x$ 
Let x2 - 1 = t
$\Rightarrow$ (2x)dx = dt
$\Rightarrow I_1 = \frac{1}{2} \int \frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}=\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\frac{1}{2}[2 \sqrt{\mathrm{t}}]$ = $\frac{1}{2}[2 \sqrt{\mathrm{x^2-1}}]$ 
And $2 \int \frac{1}{\sqrt{x^{2}-1}} d x=2 \log |x+\sqrt{x^{2}-1}|$ 
$\Rightarrow \int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}=\sqrt{\mathrm{x}^{2}-1}+2 \log |\mathrm{x}+\sqrt{\mathrm{x}^{2}-1}|+\mathrm{C}$

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