integral of (x-3)/((x+2)(x+1)^2)

解答

\int \frac{x - 3}{( x + 2 ) ( x + 1 ) ^{2}} d x

\frac{4}{x + 1} - 5 ln ∣ x + 2 ∣ + 5 ln ∣ x + 1 ∣ + C

求解步骤

\int \frac{x - 3}{( x + 2 ) ( x + 1 ) ^{2}} d x

乘开

\frac{x - 3}{( x + 2 ) ( x + 1 ) ^{2}} : \frac{x}{( x + 2 ) ( x + 1 ) ^{2}} - \frac{3}{( x + 2 ) ( x + 1 ) ^{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{x}{( x + 2 ) ( x + 1 ) ^{2}} d x - \int \frac{3}{( x + 2 ) ( x + 1 ) ^{2}} d x

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

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

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

化简

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

= \frac{4}{x + 1} - 5 ln ∣ x + 2 ∣ + 5 ln ∣ x + 1 ∣

解答补常数

= \frac{4}{x + 1} - 5 ln ∣ x + 2 ∣ + 5 ln ∣ x + 1 ∣ + C

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