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Integrate the function
$\frac{x+2}{\sqrt{x^2-1}}$
$\frac{x+2}{\sqrt{x^2-1}}$
Solution:
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Verified Answer
$\int \frac{x+2}{\sqrt{x^2-1}} d x=\int \frac{x}{\sqrt{x^2-1}} d x+\int \frac{2}{\sqrt{x^2-1}} d x$
$=\mathrm{I}_1+\mathrm{I}_2+\mathrm{C}$ (say)
Put $x^2-1=t, \Rightarrow 2 x d x=d t$
$\mathrm{I}_1=\int \frac{x}{x^2-1} d x=\frac{1}{2} \int \frac{d t}{\sqrt{t}}=\sqrt{t}=\sqrt{x^2-1}$
and $I_2=\int \frac{2}{\sqrt{x^2-1}} d x=2 \log \mid x+\sqrt{x^2-1 \mid}$
Hence $\mathrm{I}=\sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2-1}\right|+C$
$=\mathrm{I}_1+\mathrm{I}_2+\mathrm{C}$ (say)
Put $x^2-1=t, \Rightarrow 2 x d x=d t$
$\mathrm{I}_1=\int \frac{x}{x^2-1} d x=\frac{1}{2} \int \frac{d t}{\sqrt{t}}=\sqrt{t}=\sqrt{x^2-1}$
and $I_2=\int \frac{2}{\sqrt{x^2-1}} d x=2 \log \mid x+\sqrt{x^2-1 \mid}$
Hence $\mathrm{I}=\sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2-1}\right|+C$
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