Question

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∫dx/ корень 2x^2-x+3

Answer 1

$\int \frac{dx}{2x^2-x+3}=\Big [\; 2x^2-x+3=2(x^2-\frac{1}{2}x+\frac{3}{2})=2\cdot ((x-\frac{1}{4})^2-\frac{1}{16}+\frac{3}{2})=\\\\=2\cdot ((x-\frac{1}{4})^2+\frac{23}{8})\; \Big ]=\frac{1}{2}\int \frac{dx}{(x-\frac{1}{4})^2+\frac{23}{8}}=\frac{1}{2}\int \frac{d(x-\frac{1}{4})}{(x-\frac{1}{4})^2+\frac{23}{8}}=\\\\=\frac{1}{2}\cdot \frac{2\sqrt2}{\sqrt{23}}\cdot arctg\frac{2\sqrt2\, (x-\frac{1}{4})}{\sqrt{23}}+C=\sqrt{\frac{2}{23}}\cdot arctg\frac{\sqrt2\, (4x-1)}{2\sqrt{23}}+C$