Question
$\int \frac{1}{\sqrt{x^2-8 x-20}} d x$
✓
Answer
$\text { Let } I =\int \frac{1}{\sqrt{x^2-8 x-20}} d x$
$=\int \frac{1}{\sqrt{x^2-2.4 x+16-16-20}} d x$
$=\int \frac{ d x}{\sqrt{(x-4)^2-36}} d x$
$=\int \frac{ d x}{\sqrt{(x-4)^2-6^2}} d x$
$=\log \left|(x-4)+\sqrt{(x-4)^2-6^2}\right|+ c$
$\therefore I =\log \left|(x-4)+\sqrt{x^2-8 x-20}\right|+ c $
$=\int \frac{1}{\sqrt{x^2-2.4 x+16-16-20}} d x$
$=\int \frac{ d x}{\sqrt{(x-4)^2-36}} d x$
$=\int \frac{ d x}{\sqrt{(x-4)^2-6^2}} d x$
$=\log \left|(x-4)+\sqrt{(x-4)^2-6^2}\right|+ c$
$\therefore I =\log \left|(x-4)+\sqrt{x^2-8 x-20}\right|+ c $
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