Integral of (sqrt(2x^2-3))/x^2 Can someone please help me with this problem? I've stared at it and tried to figure it out for half an hour now to no avail. Thanks.
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Are you trying to say $\displaystyle \int \frac{\sqrt{2x^{2 - 3}}}{x^2}$ or$\displaystyle \int \sqrt \frac{{2x^{2 - 3}}}{x^2}$
Originally Posted by virtuoso735 Integral of (sqrt(2x^2-3))/x^2 Can someone please help me with this problem? I've stared at it and tried to figure it out for half an hour now to no avail. Thanks. Try trig substitution: Let $\displaystyle x=\frac{1}{\sqrt{6}} \sec \theta$, then $\displaystyle 6x^2-1 =\sec^2 \theta -1= \tan^2 \theta$ Then $\displaystyle \int \frac{\sqrt{2x^2-3}}{x^2}= \int \frac{\sqrt{\frac{1}{3}(6x^2-1)}}{x^2}=...$
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