integral from 1 to 4 of ((3x^2+9x+1)/x)

解答

\int_{1}^{4} (\frac{3 x ^{2} + 9 x + 1}{x}) d x

2 ln (2) + \frac{99}{2}

十进制

50.88629 \dots

求解步骤

\int_{1}^{4} \frac{3 x ^{2} + 9 x + 1}{x} d x

乘开

\frac{3 x ^{2} + 9 x + 1}{x} : 3 x + 9 + \frac{1}{x}

= \int_{1}^{4} 3 x + 9 + \frac{1}{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}^{4} 3 x d x + \int_{1}^{4} 9 d x + \int_{1}^{4} \frac{1}{x} d x

\int_{1}^{4} 3 x d x = \frac{45}{2}

\int_{1}^{4} 9 d x = 27

\int_{1}^{4} \frac{1}{x} d x = 2 ln (2)

= \frac{45}{2} + 27 + 2 ln (2)

合并分式

27 + \frac{45}{2} : \frac{99}{2}

= 2 ln (2) + \frac{99}{2}

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