integral from 1 to 2 of (x+4x^2-2)/x

解答

\int_{1}^{2} \frac{x + 4 x ^{2} - 2}{x} d x

- 2 ln (2) + 7

十进制

5.61370 \dots

求解步骤

\int_{1}^{2} \frac{x + 4 x ^{2} - 2}{x} d x

乘开

\frac{x + 4 x ^{2} - 2}{x} : 1 + 4 x - \frac{2}{x}

= \int_{1}^{2} 1 + 4 x - \frac{2}{x} d x

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}^{2} 1 d x + \int_{1}^{2} 4 x d x - \int_{1}^{2} \frac{2}{x} d x

\int_{1}^{2} 1 d x = 1

\int_{1}^{2} 4 x d x = 6

\int_{1}^{2} \frac{2}{x} d x = 2 ln (2)

= 1 + 6 - 2 ln (2)

化简

= - 2 ln (2) + 7

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