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\( \int \frac{d x}{\sqrt{x \sqrt{x}-x^{2}}} \) |
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Answer» Let I = \(\int\frac{dx}{\sqrt{x\sqrt x-x^2}}\) = \(\int\frac{dx}{x\sqrt{\frac1x-1}}\) Let \(\frac1{\sqrt x}-1=t^2\) ⇒ \(\frac1{\sqrt x}=1+t^2\) ⇒ \(\sqrt x = \frac{1}{1+t^2}\) \(\therefore\) \(-\frac12x^{-3/2}dx\) = 2tdt ⇒ \(\frac1{x\sqrt x}dx=-4t dt\) ⇒ \(\frac1xdx\) = -ut\(\sqrt x\)dt = \(\frac{-4t}{1+t^2}dt\) \(\therefore\) I = \(\int\frac{-4tdt}{(1+t^2).t}\) = -4\(\int\frac{dt}{1+t^2}\) = -4 tan-1 t + c = -4 tan-1(\(\sqrt{\frac1{\sqrt x}-1}+c\)) (\(\therefore t = \sqrt{\frac1{\sqrt x}-1}\)) \(\therefore\) \(\int\frac{dx}{\sqrt{x\sqrt x-x^2}}\) = -4 tan-1\(\sqrt{\frac1{\sqrt x}-1}+c\) |
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