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Question Number 130531 by mathmax by abdo last updated on 26/Jan/21

$$\mathrm{find}{\int }_0^1\arctan ({\mathrm{x}}^2+\mathrm{x}+1)\mathrm{dx}$$

Answered by Lordose last updated on 26/Jan/21

$$$$$$\mathrm{\Omega }={\int }_0^1{\tan }^{-1}({\mathrm{x}}^2+\mathrm{x}+1)\stackrel{\mathrm{IBP}}{=}{\mathrm{xtan}}^{-1}({\mathrm{x}}^2+\mathrm{x}+1){\mid }_0^1-{\int }_0^1\frac{\mathrm{x}(2\mathrm{x}+1)}{1+{({\mathrm{x}}^2+\mathrm{x}+1)}^2}\mathrm{dx}$$$$\mathrm{\Omega }={\tan }^{-1}\left(3\right)-\mathrm{\Phi }$$$$\mathrm{\Phi }={\int }_0^1(\frac{\mathrm{x}}{{\mathrm{x}}^2+1}-\frac{\mathrm{x}}{{\mathrm{x}}^2+2\mathrm{x}+2})\mathrm{dx}=\frac{1}{2}\log ({\mathrm{x}}^2+1){\mid }_0^1-(\frac{1}{2}{\int }_0^1\frac{2\mathrm{x}+2}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}-{\int }_0^1\frac{1}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx})$$$$\mathrm{\Phi }=\frac{1}{2}\log \left(2\right)-\frac{1}{2}\mathrm{\Psi }+\mathrm{\Pi }$$$$\mathrm{\Psi }={\int }_0^1\frac{2\mathrm{x}+2}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}\stackrel{\mathrm{u}={\mathrm{x}}^2+2\mathrm{x}+2}{=}{\int }_2^5\frac{1}{\mathrm{u}}\mathrm{du}=\log \left(\frac{5}{2}\right)$$$$\mathrm{\Pi }={\int }_0^1\frac{1}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}={\int }_0^1\frac{1}{1+{(\mathrm{x}+1)}^2}\mathrm{dx}\stackrel{\mathrm{u}=\mathrm{x}+1}{=}{\int }_1^2\frac{1}{1+{\mathrm{u}}^2}\mathrm{du}={\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)$$$$\mathrm{\Phi }=\frac{1}{2}\log \left(2\right)-\frac{1}{2}\log \left(\frac{5}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)$$$$\mathrm{\Omega }={\tan }^{-1}\left(3\right)-\frac{1}{2}\log \left(2\right)+\frac{1}{2}\log \left(\frac{5}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)$$$$\mathrm{\Omega }={\tan }^{-1}\left(\frac{4}{3}\right)-\frac{1}{2}(\log \left(2\right)-\log \left(\frac{5}{2}\right))$$

Commented by mathmax by abdo last updated on 26/Jan/21

$$\mathrm{from}\mathrm{where}\mathrm{come}\mathrm{the}\mathrm{decomposition}\frac{\mathrm{x}}{{\mathrm{x}}^2+1}-\frac{\mathrm{x}}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{sir}?$$

Commented by Lordose last updated on 27/Jan/21

$$\mathrm{Partial}\mathrm{fraction}\mathrm{decomposition}$$

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