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 $x^2 - 1 = t$
$\Rightarrow$ (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 $2 \int \frac{1}{\sqrt{x^{2}-1}} d x=2 \log |x+\sqrt{x^{2}-1}|$
$\Rightarrow \int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}=\sqrt{\mathrm{x}^{2}-1}+2 \log |\mathrm{x}+\sqrt{\mathrm{x}^{2}-1}|+\mathrm{C}$
$\Rightarrow \int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$
$= \frac{1}{2} \int \frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}+2 \int \frac{1}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{d} \mathrm{x}$
Now in, $I_1$= $\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x$
Let $x^2 - 1 = t$
$\Rightarrow$ (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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