integral of ((3-x))/(16+6x-x^2)

解答

\int \frac{( 3 - x )}{16 + 6 x - x ^{2}} d x

\frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x - 8 ∣ - 1 + C

求解步骤

\int \frac{3 - x}{16 + 6 x - x ^{2}} d x

乘开

\frac{3 - x}{16 + 6 x - x ^{2}} : \frac{3}{16 + 6 x - x ^{2}} - \frac{x}{16 + 6 x - x ^{2}}

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{3}{16 + 6 x - x ^{2}} d x - \int \frac{x}{16 + 6 x - x ^{2}} d x

\int \frac{3}{16 + 6 x - x ^{2}} d x = 3 (\frac{1}{10} ln ∣ x + 2 ∣ - \frac{1}{10} ln ∣ x - 8 ∣)

\int \frac{x}{16 + 6 x - x ^{2}} d x = - \frac{1}{5} ln ∣ x + 2 ∣ - \frac{4}{5} ln ∣ x - 8 ∣ + 1

= 3 (\frac{1}{10} ln ∣ x + 2 ∣ - \frac{1}{10} ln ∣ x - 8 ∣) - (- \frac{1}{5} ln ∣ x + 2 ∣ - \frac{4}{5} ln ∣ x - 8 ∣ + 1)

化简

3 (\frac{1}{10} ln ∣ x + 2 ∣ - \frac{1}{10} ln ∣ x - 8 ∣) - (- \frac{1}{5} ln ∣ x + 2 ∣ - \frac{4}{5} ln ∣ x - 8 ∣ + 1) : \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x - 8 ∣ - 1

= \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x - 8 ∣ - 1

解答补常数

= \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x - 8 ∣ - 1 + C

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