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受欢迎的微积分 >

integral of ((x+1))/(x^2+5x+6)

解答

\int \frac{( x + 1 )}{x ^{2} + 5 x + 6} d x

解答

2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 + \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣)) + C

求解步骤

\int \frac{x + 1}{x ^{2} + 5 x + 6} d x

对

x^{2} + 5 x + 6

配方:

(x + \frac{5}{2})^{2} - \frac{1}{4}

= \int \frac{x + 1}{( x + \frac{5}{2} ) ^{2} - \frac{1}{4}} d x

使用换元积分法

= \int \frac{2 ( 2 u - 3 )}{4 u ^{2} - 1} d u

提出常数:

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

= 2 \cdot \int \frac{2 u - 3}{4 u ^{2} - 1} d u

乘开

\frac{2 u - 3}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} - \frac{3}{4 u ^{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

= 2 (\int \frac{2 u}{4 u ^{2} - 1} d u - \int \frac{3}{4 u ^{2} - 1} d u)

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

\int \frac{3}{4 u ^{2} - 1} d u = - \frac{3}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 2 (\frac{1}{4} ln 4 u^{2} - 1 - (- \frac{3}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)))

u = x + \frac{5}{2}

代回

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

化简

2 (\frac{1}{4} ln 4 (x + \frac{5}{2})^{2} - 1 - (- \frac{3}{4} (ln 2 (x + \frac{5}{2}) + 1 - ln 2 (x + \frac{5}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 + \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 + \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣))

解答补常数

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 + \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣)) + C

作图

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