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受欢迎的微积分 >

integral of ((x+5)/(x^2+11x+18))

解答

\int (\frac{x + 5}{x ^{2} + 11 x + 18}) d x

解答

2 (\frac{1}{4} ln 4 x^{2} + 44 x + 72 + \frac{1}{28} (ln \frac{2 x + 11}{7} + 1 - ln \frac{2 x + 11}{7} - 1)) + C

求解步骤

\int \frac{x + 5}{x ^{2} + 11 x + 18} d x

使用换元积分法

= \int \frac{u}{u ^{2} + u - 12} d u

对

u^{2} + u - 12

配方:

(u + \frac{1}{2})^{2} - \frac{49}{4}

= \int \frac{u}{( u + \frac{1}{2} ) ^{2} - \frac{49}{4}} d u

使用换元积分法

= \int \frac{2 ( 2 v - 1 )}{4 v ^{2} - 49} d v

提出常数:

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

= 2 \cdot \int \frac{2 v - 1}{4 v ^{2} - 49} d v

乘开

\frac{2 v - 1}{4 v ^{2} - 49} : \frac{2 v}{4 v ^{2} - 49} - \frac{1}{4 v ^{2} - 49}

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

= 2 (\int \frac{2 v}{4 v ^{2} - 49} d v - \int \frac{1}{4 v ^{2} - 49} d v)

\int \frac{2 v}{4 v ^{2} - 49} d v = \frac{1}{4} ln 4 v^{2} - 49

\int \frac{1}{4 v ^{2} - 49} d v = - \frac{1}{28} (ln \frac{2}{7} v + 1 - ln \frac{2}{7} v - 1)

= 2 (\frac{1}{4} ln 4 v^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} v + 1 - ln \frac{2}{7} v - 1)))

代回

= 2 (\frac{1}{4} ln 4 (x + 5 + \frac{1}{2})^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} (x + 5 + \frac{1}{2}) + 1 - ln \frac{2}{7} (x + 5 + \frac{1}{2}) - 1)))

化简

2 (\frac{1}{4} ln 4 (x + 5 + \frac{1}{2})^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} (x + 5 + \frac{1}{2}) + 1 - ln \frac{2}{7} (x + 5 + \frac{1}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 44 x + 72 + \frac{1}{28} (ln \frac{2 x + 11}{7} + 1 - ln \frac{2 x + 11}{7} - 1))

= 2 (\frac{1}{4} ln 4 x^{2} + 44 x + 72 + \frac{1}{28} (ln \frac{2 x + 11}{7} + 1 - ln \frac{2 x + 11}{7} - 1))

解答补常数

= 2 (\frac{1}{4} ln 4 x^{2} + 44 x + 72 + \frac{1}{28} (ln \frac{2 x + 11}{7} + 1 - ln \frac{2 x + 11}{7} - 1)) + C

作图

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