Question

Solve: ${\int }_0^1\frac{1}{{2x}^2+x+1}dx$

Answer 1

$\int \frac{dx}{2x^2+x+1}=\frac{1}{2}\int \frac{dx}{x^2+\frac{x}{2}+\frac{1}{2}}$

=$\frac{1}{2}\int \frac{dx}{x^2+\frac{x}{2}+\frac{1}{16}-\frac{1}{16}+\frac{1}{12}}$

=$\frac{1}{2}\int \frac{dx}{(x+\frac{1}{4}{)}^2+(\frac{\sqrt{7}}{4}{)}^2}$

$\because \int \frac{dx}{x^2+a^2}=\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)+c$

$\because \frac{1}{2}\times \frac{1\times 4}{\sqrt{7}}\left[ta{n}^{-1}\right(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}}){]}^1$

$=\frac{2}{\sqrt{7}}\left[ta{n}^{-1}\right(\frac{5}{\sqrt{7}})-1ta{n}^{-1}(\frac{1}{\sqrt{7}}\left)\right]$

$=\frac{2}{\sqrt{7}}ta{n}^{-1}\left(\frac{\frac{5}{\sqrt{7}}-\frac{1}{\sqrt{7}}}{1+\frac{5\times 1}{\sqrt{7}\times \sqrt{7}}}\right)$

$=\frac{2}{\sqrt{7}}ta{n}^{-1}\left(\frac{4\sqrt{7}}{12}\right)$

$=\frac{2}{\sqrt{7}}ta{n}^{-1}\left(\frac{\sqrt{7}}{3}\right)$