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受欢迎的微积分 >

integral of (x+4)/(x^2-5x+6)

解答

\int \frac{x + 4}{x ^{2} - 5 x + 6} d x

解答

2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{13}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + C

求解步骤

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

使用换元积分法

= \int \frac{2 ( 2 u + 13 )}{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 + 13}{4 u ^{2} - 1} d u

乘开

\frac{2 u + 13}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} + \frac{13}{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{13}{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{13}{4 u ^{2} - 1} d u = - \frac{13}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 2 (\frac{1}{4} ln 4 u^{2} - 1 - \frac{13}{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{13}{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{13}{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{13}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{13}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣))

解答补常数

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{13}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + C

作图

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