1 3 x^2+8 d x - Vidyadip

Question

$\int \frac{1}{\sqrt{3 x^2+8}} d x$

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Answer

$ \int \frac{1}{\sqrt{3 x^2+8}} d x=\frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{x^2+\left(\frac{2 \sqrt{2}}{\sqrt{3}}\right)^2}} d x$
$=\frac{1}{\sqrt{3}} \log \left|x+\sqrt{x^2+\left(2 \frac{\sqrt{2}}{\sqrt{3}}\right)^2}\right|+ c _1$
$=\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{3} x+\sqrt{3 x^2+8}}{\sqrt{3}}\right|+ c _1$
$=\frac{1}{\sqrt{3}} \log \left|\sqrt{3} x+\sqrt{3 x^2+8}\right|+ c _{,} $
where $c = c _1-\frac{1}{\sqrt{3}} \log \sqrt{3}$

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