integral of (x+2)/(2x^2+x+1)

解答

\int \frac{x + 2}{2 x ^{2} + x + 1} d x

\frac{1}{4} ln 16 x^{2} + 8 x + 8 + \frac{7}{2} arctan (\frac{1}{7} (4 x + 1)) + C

求解步骤

\int \frac{x + 2}{2 x ^{2} + x + 1} d x

乘开

\frac{x + 2}{2 x ^{2} + x + 1} : \frac{x}{2 x ^{2} + x + 1} + \frac{2}{2 x ^{2} + x + 1}

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{2 x ^{2} + x + 1} d x + \int \frac{2}{2 x ^{2} + x + 1} d x

\int \frac{x}{2 x ^{2} + x + 1} d x = 2 (\frac{1}{8} ln 16 x^{2} + 8 x + 8 - \frac{1}{4 7} arctan (\frac{4 x + 1}{7}))

\int \frac{2}{2 x ^{2} + x + 1} d x = \frac{4}{7} arctan (\frac{4 x + 1}{7})

= 2 (\frac{1}{8} ln 16 x^{2} + 8 x + 8 - \frac{1}{4 7} arctan (\frac{4 x + 1}{7})) + \frac{4}{7} arctan (\frac{4 x + 1}{7})

化简

2 (\frac{1}{8} ln 16 x^{2} + 8 x + 8 - \frac{1}{4 7} arctan (\frac{4 x + 1}{7})) + \frac{4}{7} arctan (\frac{4 x + 1}{7}) : \frac{1}{4} ln 16 x^{2} + 8 x + 8 + \frac{7}{2} arctan (\frac{1}{7} (4 x + 1))

= \frac{1}{4} ln 16 x^{2} + 8 x + 8 + \frac{7}{2} arctan (\frac{1}{7} (4 x + 1))

解答补常数

= \frac{1}{4} ln 16 x^{2} + 8 x + 8 + \frac{7}{2} arctan (\frac{1}{7} (4 x + 1)) + C

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