integral from 1 to 2 of (13)/(x^3+4x)

解答

\int_{1}^{2} \frac{13}{x ^{3} + 4 x} d x

13 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

十进制

1.48897 \dots

求解步骤

\int_{1}^{2} \frac{13}{x ^{3} + 4 x} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 13 \cdot \int_{1}^{2} \frac{1}{x ^{3} + 4 x} d x

将

\frac{1}{x ^{3} + 4 x}

用部份分式展开:

\frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}

= 13 \cdot \int_{1}^{2} \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )} 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

= 13 (\int_{1}^{2} \frac{1}{4 x} d x - \int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x)

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

\int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} (3 ln (2) - ln (5))

= 13 (\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)))

化简

\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)) : \frac{- ln ( 2 ) + ln ( 5 )}{8}

= 13 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

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