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受欢迎的微积分 >

integral of (3x-2)/(x^2+2x+17)

解答

\int \frac{3 x - 2}{x ^{2} + 2 x + 17} d x

解答

\frac{1}{4} (6 ln \frac{1}{16} (x + 1)^{2} + 1 - 5 arctan (\frac{1}{4} (x + 1))) + C

求解步骤

\int \frac{3 x - 2}{x ^{2} + 2 x + 17} d x

对

x^{2} + 2 x + 17

配方:

(x + 1)^{2} + 16

= \int \frac{3 x - 2}{( x + 1 ) ^{2} + 16} d x

使用换元积分法

= \int \frac{3 u - 5}{u ^{2} + 16} d u

使用换元积分法

= \int \frac{12 v - 5}{4 ( 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}{4} \cdot \int \frac{12 v - 5}{v ^{2} + 1} d v

乘开

\frac{12 v - 5}{v ^{2} + 1} : \frac{12 v}{v ^{2} + 1} - \frac{5}{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}{4} (\int \frac{12 v}{v ^{2} + 1} d v - \int \frac{5}{v ^{2} + 1} d v)

\int \frac{12 v}{v ^{2} + 1} d v = 6 ln v^{2} + 1

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

= \frac{1}{4} (6 ln v^{2} + 1 - 5 arctan (v))

代回

= \frac{1}{4} (6 ln (\frac{x + 1}{4})^{2} + 1 - 5 arctan (\frac{x + 1}{4}))

化简

\frac{1}{4} (6 ln (\frac{x + 1}{4})^{2} + 1 - 5 arctan (\frac{x + 1}{4})) : \frac{1}{4} (6 ln \frac{1}{16} (x + 1)^{2} + 1 - 5 arctan (\frac{1}{4} (x + 1)))

= \frac{1}{4} (6 ln \frac{1}{16} (x + 1)^{2} + 1 - 5 arctan (\frac{1}{4} (x + 1)))

解答补常数

= \frac{1}{4} (6 ln \frac{1}{16} (x + 1)^{2} + 1 - 5 arctan (\frac{1}{4} (x + 1))) + C

作图

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