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Question Number 137477 by Sandeep11 last updated on 03/Apr/21

Commented by MJS_new last updated on 03/Apr/21

$$\mathrm{well}\mathrm{just}\mathrm{derivate}\mathrm{the}\mathrm{answers}\mathrm{and}\mathrm{find}\mathrm{the}$$$$\mathrm{right}\mathrm{one}?!$$

Answered by soumyasaha last updated on 03/Apr/21

$$=\int \frac{{\mathrm{x}}^2(1-\frac{1}{{\mathrm{x}}^2})}{\mathrm{x}.\mathrm{x}\sqrt{(\mathrm{x}+\frac{1}{\mathrm{x}}+\alpha )(\mathrm{x}+\frac{1}{\mathrm{x}}+\beta )}}\mathrm{dx}$$$$=\int \frac{(1-\frac{1}{{\mathrm{x}}^2})}{\sqrt{(\mathrm{x}+\frac{1}{\mathrm{x}}+\alpha )(\mathrm{x}+\frac{1}{\mathrm{x}}+\beta )}}\mathrm{dx}$$$$\mathrm{Let},\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow (1-\frac{1}{{\mathrm{x}}^2})\mathrm{dx}=\mathrm{dt}$$$$\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{\sqrt{(\mathrm{t}+\alpha )(\mathrm{t}+\beta )}}$$$$\mathrm{Let},\sqrt{\mathrm{t}+\alpha }=\mathrm{z}\Rightarrow \mathrm{t}+\alpha ={\mathrm{z}}^2\Rightarrow \mathrm{dt}=2\mathrm{zdz}$$$$\therefore \mathrm{I}=\int \frac{2\mathrm{zdz}}{\sqrt{{\mathrm{z}}^2({\mathrm{z}}^2+\beta -\alpha )}}$$$$=2.\int \frac{\mathrm{dz}}{\sqrt{{\mathrm{z}}^2+(\beta -\alpha )}}$$$$=2\ln \mid \mathrm{z}+\sqrt{{\mathrm{z}}^2+\beta -\alpha }\mid +\mathrm{C}$$$$=2\ln \mid \sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}+\alpha }+\sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}+\beta }\mid +\mathrm{C}$$$$=2\ln \mid \frac{\sqrt{{\mathrm{x}}^2+\alpha \mathrm{x}+1}+\sqrt{{\mathrm{x}}^2+\beta \mathrm{x}+1}}{\sqrt{\mathrm{x}}}\mid +\mathrm{C}$$

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