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受欢迎的微积分 >

integral from 6 to 8 of x/(x^2+4x+20)

解答

\int_{6}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

解答

\frac{1}{2} (ln (\frac{29}{4}) - ln (5) - arctan (\frac{5}{2}) + arctan (2))

+1

十进制

0.14421 \dots

求解步骤

\int_{6}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

对

x^{2} + 4 x + 20

配方:

(x + 2)^{2} + 16

= \int_{6}^{8} \frac{x}{( x + 2 ) ^{2} + 16} d x

使用换元积分法

= \int_{8}^{10} \frac{u - 2}{u ^{2} + 16} d u

使用换元积分法

= \int_{2}^{\frac{5}{2}} \frac{2 v - 1}{2 ( v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{2} \cdot \int_{2}^{\frac{5}{2}} \frac{2 v - 1}{v ^{2} + 1} d v

乘开

\frac{2 v - 1}{v ^{2} + 1} : \frac{2 v}{v ^{2} + 1} - \frac{1}{v ^{2} + 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

= \frac{1}{2} (\int_{2}^{\frac{5}{2}} \frac{2 v}{v ^{2} + 1} d v - \int_{2}^{\frac{5}{2}} \frac{1}{v ^{2} + 1} d v)

\int_{2}^{\frac{5}{2}} \frac{2 v}{v ^{2} + 1} d v = ln (\frac{29}{4}) - ln (5)

\int_{2}^{\frac{5}{2}} \frac{1}{v ^{2} + 1} d v = arctan (\frac{5}{2}) - arctan (2)

= \frac{1}{2} (ln (\frac{29}{4}) - ln (5) - (arctan (\frac{5}{2}) - arctan (2)))

化简

= \frac{1}{2} (ln (\frac{29}{4}) - ln (5) - arctan (\frac{5}{2}) + arctan (2))

作图

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