integral from 1 to 3 of (2x+1)/(x^2+x-1)

解答

\int_{1}^{3} \frac{2 x + 1}{x ^{2} + x - 1} d x

ln (11)

十进制

2.39789 \dots

求解步骤

\int_{1}^{3} \frac{2 x + 1}{x ^{2} + x - 1} d x

乘开

\frac{2 x + 1}{x ^{2} + x - 1} : \frac{2 x}{x ^{2} + x - 1} + \frac{1}{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_{1}^{3} \frac{2 x}{x ^{2} + x - 1} d x + \int_{1}^{3} \frac{1}{x ^{2} + x - 1} d x

\int_{1}^{3} \frac{2 x}{x ^{2} + x - 1} d x = ln (11) + \frac{1}{5} ln (\frac{21 - 8 5}{11})

\int_{1}^{3} \frac{1}{x ^{2} + x - 1} d x = - \frac{ln ( \frac{21 - 8 5}{11} )}{5}

= ln (11) + \frac{1}{5} ln (\frac{21 - 8 5}{11}) - \frac{ln ( \frac{21 - 8 5}{11} )}{5}

化简

ln (11) + \frac{1}{5} ln (\frac{21 - 8 5}{11}) - \frac{ln ( \frac{21 - 8 5}{11} )}{5} : ln (11)

= ln (11)

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