integral of (2x-16sqrt(x))/(x^2-7x)

解答

\int \frac{2 x - 16 x}{x ^{2} - 7 x} d x

2 ln ∣ x - 7 ∣ + \frac{16}{7} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1) + C

求解步骤

\int \frac{2 x - 16 x}{x ^{2} - 7 x} d x

乘开

\frac{2 x - 16 x}{x ^{2} - 7 x} : \frac{2 x}{x ^{2} - 7 x} - \frac{16 x}{x ^{2} - 7 x}

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{2 x}{x ^{2} - 7 x} d x - \int \frac{16 x}{x ^{2} - 7 x} d x

\int \frac{2 x}{x ^{2} - 7 x} d x = 2 ln ∣ x - 7 ∣

\int \frac{16 x}{x ^{2} - 7 x} d x = - \frac{16}{7} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1)

= 2 ln ∣ x - 7 ∣ - (- \frac{16}{7} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1))

化简

= 2 ln ∣ x - 7 ∣ + \frac{16}{7} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1)

解答补常数

= 2 ln ∣ x - 7 ∣ + \frac{16}{7} (ln \frac{x}{7} + 1 - ln \frac{x}{7} - 1) + C

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