integral of (6x^2+13x+6)/((x+2)(x+1)^2)

解答

\int \frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} d x

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

求解步骤

\int \frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} d x

将

\frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}}

用部份分式展开:

\frac{4}{x + 2} + \frac{2}{x + 1} - \frac{1}{( x + 1 ) ^{2}}

= \int \frac{4}{x + 2} + \frac{2}{x + 1} - \frac{1}{( x + 1 ) ^{2}} d 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{4}{x + 2} d x + \int \frac{2}{x + 1} d x - \int \frac{1}{( x + 1 ) ^{2}} d x

\int \frac{4}{x + 2} d x = 4 ln ∣ x + 2 ∣

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

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

化简

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

解答补常数

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

t an g e n t y = e^{x^{3}}, (- 1, \frac{1}{e})

d er i v a t i v e \frac{( y - 2 ) ^{3}}{( y ^{2} + 4 y ) ^{7}}

x \to + 3 lim (f (x))

x \to 90 lim (sin (x))

\frac{d y}{d x} = \frac{x}{16 y}