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Question Number 121859 by Bird last updated on 12/Nov/20

$$find\int \frac{dx}{(x-1)\sqrt{x+1}-(x+1)\sqrt{x-1}}$$

Answered by ajfour last updated on 12/Nov/20

$$I=\int \frac{\frac{dx}{\sqrt{x^2-1}}}{\sqrt{x-1}-\sqrt{x+1}}$$$$$$$$=\int \frac{(\sqrt{x-1}+\sqrt{x+1})dx}{2\sqrt{x^2-1}}$$$$2I=\int \frac{dx}{\sqrt{x+1}}+\int \frac{dx}{\sqrt{x-1}}$$$$I=\sqrt{x+1}+\sqrt{x-1}+c$$$$$$

Answered by liberty last updated on 12/Nov/20

$$\frac{1}{(\mathrm{x}-1)\sqrt{\mathrm{x}+1}-(\mathrm{x}+1)\sqrt{\mathrm{x}-1}}.\frac{(\mathrm{x}-1)\sqrt{\mathrm{x}+1}+(\mathrm{x}+1)\sqrt{\mathrm{x}-1}}{(\mathrm{x}-1)\sqrt{\mathrm{x}+1}+(\mathrm{x}+1)\sqrt{\mathrm{x}-1}}=$$$$\frac{(\mathrm{x}-1)\sqrt{\mathrm{x}+1}+(\mathrm{x}+1)\sqrt{\mathrm{x}-1}}{{(\mathrm{x}-1)}^2(\mathrm{x}+1)+({\mathrm{x}}^2-1)\sqrt{{\mathrm{x}}^2-1}-({\mathrm{x}}^2-1)\sqrt{{\mathrm{x}}^2-1}-{(\mathrm{x}+1)}^2(\mathrm{x}-1)}=$$$$\frac{(\mathrm{x}-1)\sqrt{\mathrm{x}+1}+(\mathrm{x}+1)\sqrt{\mathrm{x}-1}}{{(\mathrm{x}-1)}^2(\mathrm{x}+1)-{(\mathrm{x}+1)}^2(\mathrm{x}-1)}=$$$$\frac{(\mathrm{x}-1)\sqrt{\mathrm{x}+1}+(\mathrm{x}+1)\sqrt{\mathrm{x}-1}}{({\mathrm{x}}^2-1)\{\mathrm{x}-1-\mathrm{x}-1\}}=$$$$\frac{(\mathrm{x}-1)\sqrt{\mathrm{x}+1}+(\mathrm{x}+1)\sqrt{\mathrm{x}-1}}{-2({\mathrm{x}}^2-1)}=$$$$-\frac{1}{2}[\frac{\sqrt{\mathrm{x}+1}}{\mathrm{x}+1}+\frac{\sqrt{\mathrm{x}-1}}{\mathrm{x}-1}]=-\frac{1}{2}[\frac{1}{\sqrt{\mathrm{x}+1}}+\frac{1}{\sqrt{\mathrm{x}-1}}]$$$$\mathrm{then}\mathrm{I}=-\frac{1}{2}\int [{(\mathrm{x}+1)}^{-\frac{1}{2}}+{(\mathrm{x}-1)}^{-\frac{1}{2}}]\mathrm{dx}$$$$\mathrm{I}=-\frac{1}{2}[2\sqrt{\mathrm{x}+1}+2\sqrt{\mathrm{x}-1}]+\mathrm{c}$$$$\mathrm{I}=-\sqrt{\mathrm{x}+1}-\sqrt{\mathrm{x}-1}+\mathrm{c}.▴$$

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