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Question Number 128052 by bramlexs22 last updated on 04/Jan/21

$$\mathrm{I}=\int \arctan \left(\frac{\mathrm{x}-2}{\mathrm{x}+2}\right)\mathrm{dx}$$

Answered by liberty last updated on 04/Jan/21

$$\mathrm{I}=\mathrm{x}\arctan \left(\frac{\mathrm{x}-2}{\mathrm{x}+2}\right)-\int \mathrm{x}(\frac{1}{1+{\left(\frac{\mathrm{x}-2}{\mathrm{x}+2}\right)}^2}.\frac{4}{{(\mathrm{x}+2)}^2})\mathrm{dx}$$$$\mathrm{I}=\mathrm{x}\arctan \left(\frac{\mathrm{x}-2}{\mathrm{x}+2}\right)-\int \frac{2\mathrm{x}}{{\mathrm{x}}^2+4}\mathrm{dx}$$$$\mathrm{I}=\mathrm{x}\arctan \left(\frac{\mathrm{x}-2}{\mathrm{x}+2}\right)-\ln ({\mathrm{x}}^2+4)+\mathrm{c}$$

Answered by Olaf last updated on 04/Jan/21

$$\mathrm{I}=\int \arctan \left(\frac{x-2}{x+2}\right)dx$$$$\mathrm{I}=-\int \arctan \left(\frac{1-\frac{x}{2}}{1+\frac{x}{2}}\right)dx\left(1\right)$$$$\mathrm{I}=-\int [\arctan \left(1\right)-\arctan \left(\frac{x}{2}\right)]dx\left(2\right)$$$$\mathrm{I}=-\frac{\pi }{4}x+x\arctan \left(\frac{x}{2}\right)-\int \frac{2x}{x^2+4}dx$$$$\mathrm{I}=-\frac{\pi }{4}x+x\arctan \left(\frac{x}{2}\right)-\ln (x^2+4)+\mathrm{C}$$$$$$$$\left(1\right):\mathrm{artan}(-u)=-\arctan u$$$$\left(2\right)\arctan u-\arctan v=\arctan \left(\frac{u-v}{1+uv}\right)$$

Answered by bramlexs22 last updated on 04/Jan/21

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