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Question Number 85839 by sahnaz last updated on 25/Mar/20

$$\int \mathrm{x}\times \frac{1}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}$$

Commented by jagoll last updated on 25/Mar/20

$$\int \frac{\mathrm{x}\mathrm{dx}}{\sqrt{{\mathrm{x}}^2-1}}=\frac{1}{2}\int \frac{\mathrm{d}({\mathrm{x}}^2-1)}{{({\mathrm{x}}^2-1)}^{1/2}}$$$$=\frac{1}{2}\times \frac{2}{1}\times \sqrt{{\mathrm{x}}^2-1}+\mathrm{c}$$$$=\sqrt{{\mathrm{x}}^2-1}+\mathrm{c}$$

Answered by sakeefhasan05@gmail.com last updated on 25/Mar/20

$$\mathrm{let}{\mathrm{t}}^2={\mathrm{x}}^2-1$$$$\mathrm{t}.\mathrm{dt}=\mathrm{x}.\mathrm{dx}$$$$\int \frac{\mathrm{t}}{\sqrt{{\mathrm{t}}^2}}\mathrm{dt}=\int \left(1\right)\mathrm{dt}=\mathrm{t}+\mathrm{C}$$$$=\sqrt{{\mathrm{x}}^2-1}+\mathrm{C}$$$$$$$$$$

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