1.

\( \int \frac{d x}{\sqrt{x \sqrt{x}-x^{2}}} \)

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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