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Question Number 182954 by Gamil last updated on 17/Dec/22

$$\mathit{I}=\int \frac{1}{\sqrt{\mathit{x}\sqrt{\mathit{x}}-{\mathit{x}}^2}}\mathit{d}\mathit{x}$$$$\mathit{I}=\int \frac{1}{\cdot \sqrt{\mathit{x}(\sqrt{\mathit{x}}-\mathit{x})}}\mathit{d}\mathit{x}$$$$\mathit{I}=\int \frac{1}{\sqrt{\mathit{x}}\cdot \sqrt{\sqrt{\mathit{x}}-\mathit{x}}}\times \frac{2}{2}\mathit{d}\mathit{x}$$$$\mathit{I}=\int \frac{1}{\sqrt{4\sqrt{\mathit{x}}-4\mathit{x}}}\cdot \frac{2}{\sqrt{\mathit{x}}}\mathit{d}\mathit{x}$$$$\mathit{I}=2\int \frac{}{\sqrt{1-{(1-2\sqrt{\mathit{x}})}^2}}\cdot \frac{1}{\sqrt{\mathit{x}}}\mathit{d}\mathit{x}$$$$\mathit{I}=-2\int \frac{1}{\sqrt{1-{(1-2\sqrt{\mathit{x}})}^2}}\cdot \mathit{d}(1-2\sqrt{\mathit{x}})$$$$\mathit{I}=-2{\mathit{s}\mathit{i}\mathit{n}}^{-1}(1-2\sqrt{\mathit{x}})+\mathit{C}$$$$$$$$\mathbf{Gamil}\mathbf{AL}\mathbf{mansob}$$

Commented by Rasheed.Sindhi last updated on 17/Dec/22

$$$$$$\underset{\stackrel{}{\mathbf{AND}}}{\mathbf{Question}\mathrm{should}\mathrm{go}\mathrm{in}\mathrm{place}\mathrm{of}\mathbf{Question}.}$$$$\mathbf{Answer}\mathrm{should}\mathrm{go}\mathrm{in}\mathrm{place}\mathrm{of}\mathbf{Answer}.$$$$$$

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