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integral from 1 to 8 of 2log_{4}(x+3)

解答

\int_{1}^{8} 2 lo g_{4} (x + 3) d x

解答

16 lo g_{4} (11) + \frac{3 ln ( 11 ) + ln ( \frac{27}{8} ) - 3 ln ( 6 ) - 7}{ln ( 2 )} - 2

+1

十进制

19.95488 \dots

求解步骤

\int_{1}^{8} 2 lo g_{4} (x + 3) d x

提出常数:

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

= 2 \cdot \int_{1}^{8} lo g_{4} (x + 3) d x

乘以

x + 3

的共轭根式

: \frac{x ^{2} - 9}{x - 3}

= 2 \cdot \int_{1}^{8} lo g_{4} (\frac{x ^{2} - 9}{x - 3}) d x

找到无定义的点（奇点）:

x = 3

如果存在

b, a < b < c, f (b) =

无定义

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= 2 (\int_{1}^{3} lo g_{4} (\frac{x ^{2} - 9}{x - 3}) d x + \int_{3}^{8} lo g_{4} (\frac{x ^{2} - 9}{x - 3}) d x)

\int_{1}^{3} lo g_{4} (\frac{x ^{2} - 9}{x - 3}) d x = 3 lo g_{4} (6) + \frac{ln ( \frac{27}{8} ) - 2}{2 ln ( 2 )} - 1

\int_{3}^{8} lo g_{4} (\frac{x ^{2} - 9}{x - 3}) d x = - 3 lo g_{4} (6) + 8 lo g_{4} (11) + \frac{3 ln ( 11 ) - 3 ln ( 6 ) - 5}{2 ln ( 2 )}

= 2 (3 lo g_{4} (6) + \frac{ln ( \frac{27}{8} ) - 2}{2 ln ( 2 )} - 1 - 3 lo g_{4} (6) + 8 lo g_{4} (11) + \frac{3 ln ( 11 ) - 3 ln ( 6 ) - 5}{2 ln ( 2 )})

化简

2 (3 lo g_{4} (6) + \frac{ln ( \frac{27}{8} ) - 2}{2 ln ( 2 )} - 1 - 3 lo g_{4} (6) + 8 lo g_{4} (11) + \frac{3 ln ( 11 ) - 3 ln ( 6 ) - 5}{2 ln ( 2 )}) : 16 lo g_{4} (11) + \frac{3 ln ( 11 ) + ln ( \frac{27}{8} ) - 3 ln ( 6 ) - 7}{ln ( 2 )} - 2

= 16 lo g_{4} (11) + \frac{3 ln ( 11 ) + ln ( \frac{27}{8} ) - 3 ln ( 6 ) - 7}{ln ( 2 )} - 2

作图

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