integral of ((y-2)/(y^2-4y))

解答

\int (\frac{y - 2}{y ^{2} - 4 y}) d y

\frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 4 ∣ + C

求解步骤

\int \frac{y - 2}{y ^{2} - 4 y} d y

乘开

\frac{y - 2}{y ^{2} - 4 y} : \frac{y}{y ^{2} - 4 y} - \frac{2}{y ^{2} - 4 y}

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

= \int \frac{y}{y ^{2} - 4 y} d y - \int \frac{2}{y ^{2} - 4 y} d y

\int \frac{y}{y ^{2} - 4 y} d y = ln ∣ y - 4 ∣

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

= ln ∣ y - 4 ∣ - 2 (- \frac{1}{4} ln ∣ y ∣ + \frac{1}{4} ln ∣ y - 4 ∣)

化简

ln ∣ y - 4 ∣ - 2 (- \frac{1}{4} ln ∣ y ∣ + \frac{1}{4} ln ∣ y - 4 ∣) : \frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 4 ∣

= \frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 4 ∣

解答补常数

= \frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 4 ∣ + C

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