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Question Number 18278 by mondodotto@gmail.com last updated on 17/Jul/17

Commented by mondodotto@gmail.com last updated on 18/Jul/17

$$\mathrm{help}\mathrm{please}!!$$

Answered by geovane10math last updated on 18/Jul/17

$$\mathrm{Put}\mathrm{it}\mathrm{in}\mathrm{Wolfram}\mathrm{Alpha}...$$

Answered by ajfour last updated on 18/Jul/17

$$\int \frac{\left(1/{\mathrm{x}}^2\right)\mathrm{dx}}{{\mathrm{x}}^2+\left(1/{\mathrm{x}}^2\right)}=\frac{1}{2}\int \frac{(1+\frac{1}{{\mathrm{x}}^2})-(1-\frac{1}{{\mathrm{x}}^2})}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}$$$$=\int \frac{(1+\frac{1}{{\mathrm{x}}^2})\mathrm{dx}}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+2}-\int \frac{(1-\frac{1}{{\mathrm{x}}^2})\mathrm{dx}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-2}$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\mathrm{x}-1/\mathrm{x}}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\mathrm{x}+1/\mathrm{x}}{\sqrt{2}}\right)+\mathrm{C}$$$$=\frac{1}{\sqrt{2}}[{\tan }^{-1}\left(\frac{{\mathrm{x}}^2-1}{\mathrm{x}\sqrt{2}}\right)-{\tan }^{-1}\left(\frac{{\mathrm{x}}^2+1}{\mathrm{x}\sqrt{2}}\right)]+\mathrm{C}.$$

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