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Question Number 95547 by Fikret last updated on 25/May/20

$$\int \frac{2x^{3}dx}{2x^2-4x+3}=?$$

Answered by MJS last updated on 25/May/20

$$\int \frac{2x^3}{2x^2-4x+3}dx=\int (x+2+\frac{5x-6}{2x^2-4x+3})dx=$$$$=\int (x+2)dx+\frac{5}{4}\int \frac{4x-4}{2x^2-4x+3}dx-\int \frac{dx}{2x^2-4x+3}=$$$$=\frac{1}{2}{x}^2+2x+\frac{5}{4}\ln (2x^2-4x+3)-\frac{\sqrt{2}}{2}\arctan \left(\sqrt{2}(x-1)\right)+C$$

Commented by peter frank last updated on 26/May/20

$$\mathrm{thank}\mathrm{you}$$

Answered by 1549442205 last updated on 26/May/20

$$\mathrm{we}\mathrm{have}\frac{{2\mathrm{x}}^3}{{2\mathrm{x}}^2-4\mathrm{x}+3}=\mathrm{x}+2+\frac{5\mathrm{x}-6}{{2\mathrm{x}}^2-4\mathrm{x}+3}$$$$\mathrm{x}+2+\frac{5(\mathrm{x}-1)-1}{{2\mathrm{x}}^2-4\mathrm{x}+3}$$$$\mathrm{x}+2+\frac{\frac{5}{4}(4\mathrm{x}-4)-1}{{2\mathrm{x}}^2-4\mathrm{x}+3}=\mathrm{x}+2+\frac{\frac{5}{4}(4\mathrm{x}-4)}{{2\mathrm{x}}^2-4\mathrm{x}+3}$$$$-\frac{1}{{2\mathrm{x}}^2-4\mathrm{x}+3}.\mathrm{Hence},\mathrm{denoting}\mathrm{I}\mathrm{the}$$$$\mathrm{integral}\mathrm{above}\mathrm{we}\mathrm{have}:$$$$\mathrm{I}=\frac{{\mathrm{x}}^2}{2}+2\mathrm{x}+\frac{5}{4}\ln ({2\mathrm{x}}^2-4\mathrm{x}+3)-\int \frac{\mathrm{dx}}{{2\mathrm{x}}^2-4\mathrm{x}+3}$$$$\mathrm{J}=\int \frac{\mathrm{dx}}{{2\mathrm{x}}^2-4\mathrm{x}+3}=\frac{1}{2}\int \frac{\mathrm{dx}}{{(\mathrm{x}-1)}^2+{\left(\frac{\sqrt{2}}{2)}\right)}^2}$$$$=\sqrt{2}{\tan }^{-1}\left[\sqrt{2}(\mathrm{x}-1)\right]+\mathrm{C}.\mathrm{Thus},$$$$\mathrm{I}=\frac{{\mathrm{x}}^2}{2}+2\mathrm{x}+\frac{5}{4}\ln ({2\mathrm{x}}^2-4\mathrm{x}+3)-\frac{\sqrt{2}}{2}{\tan }^{-1}\left[\sqrt{2}(\mathrm{x}-1)\right]+\mathrm{C}$$$$$$

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