integral of (x+1)/((x-1)(x+2))

解答

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

\frac{2}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - 1 + C

求解步骤

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

乘以

x + 1

的共轭根式

: \frac{x ^{2} - 1}{x - 1}

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

使用分布积分法

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

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

= (x + 1) (\frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣) - \frac{1}{3} (x ln ∣ x - 1 ∣ - x ln ∣ x + 2 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 3)

化简

(x + 1) (\frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣) - \frac{1}{3} (x ln ∣ x - 1 ∣ - x ln ∣ x + 2 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 3) : \frac{2}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - 1

= \frac{2}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - 1

解答补常数

= \frac{2}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - 1 + C

\int sin (x) \cdot e^{2 x} d x

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\int 0.09 q^{2} + 4 q + 50 d q