$\displaystyle \int \frac{dx}{sqrt(x^2-4)}$
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Originally Posted by vinson24 $\displaystyle \int \frac{dx}{sqrt(x^2-4)}$ Note that $\displaystyle \int\frac{\,dx}{\sqrt{x^2-4}}=\int\frac{\,dx}{\sqrt{4\left(\frac{x^2}{4}-1\right)}}=\tfrac{1}{2}\int\frac{\,dx}{\sqrt{\left (\frac{x}{2}\right)^2-1}}$ Can you continue from here?
Originally Posted by vinson24 $\displaystyle \int \frac{dx}{\sqrt{x^2-4}}$ alternatively, a trig substitution of $\displaystyle x = 2 \sec \theta$ could work. well, not really alternatively....
yes i can thank you
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