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Question Number 125416 by bemath last updated on 10/Dec/20

$$\int \frac{dx}{\sqrt{x\sqrt{x}-x^2}}?$$

Answered by Dwaipayan Shikari last updated on 10/Dec/20

$$\int \frac{dx}{\sqrt{x}\sqrt{\sqrt{x}-x}}\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}$$$$=2\int \frac{dt}{\sqrt{t-t^2}}=2\int \frac{1}{\sqrt{t}\left(\sqrt{1-t}\right)}dtt=u^2\Rightarrow 1=2u\frac{du}{dt}$$$$=4\int \frac{du}{\sqrt{1-u^2}}=4{sin}^{-1}u+C=4{sin}^{-1}\sqrt[4]{x}dx$$

Answered by liberty last updated on 10/Dec/20

$$I=\int \frac{dx}{x\sqrt{\frac{1}{\sqrt{x}}-1}};let\frac{1}{\sqrt{x}}-1=w;x=\frac{1}{{(w+1)}^2}$$$$I=\int \frac{{(w+1)}^2}{\sqrt{w}}(-\frac{2}{{(1+w)}^3})dw$$$$I=-2\int \frac{dw}{\sqrt{w}(1+w)};let\sqrt{w}=z;dw=2zdz$$$$I=-4\int \frac{zdz}{z(1+z^2)}=-4{\tan }^{-1}\left(z\right)+c$$$$I=-4{\tan }^{-1}\left(\sqrt{w}\right)+c$$$$I=-4{\tan }^{-1}\left(\sqrt{\frac{1-\sqrt{x}}{\sqrt{x}}}\right)+c$$

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