integral of (4x^2-12x-19)/((x+1)^2(x-2))

解答

\int \frac{4 x ^{2} - 12 x - 19}{( x + 1 ) ^{2} ( x - 2 )} d x

7 ln ∣ x + 1 ∣ - \frac{1}{x + 1} - 3 ln ∣ x - 2 ∣ + C

求解步骤

\int \frac{4 x ^{2} - 12 x - 19}{( x + 1 ) ^{2} ( x - 2 )} d x

将

\frac{4 x ^{2} - 12 x - 19}{( x + 1 ) ^{2} ( x - 2 )}

用部份分式展开:

\frac{7}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - \frac{3}{x - 2}

= \int \frac{7}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - \frac{3}{x - 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{7}{x + 1} d x + \int \frac{1}{( x + 1 ) ^{2}} d x - \int \frac{3}{x - 2} d x

\int \frac{7}{x + 1} d x = 7 ln ∣ x + 1 ∣

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

\int \frac{3}{x - 2} d x = 3 ln ∣ x - 2 ∣

= 7 ln ∣ x + 1 ∣ - \frac{1}{x + 1} - 3 ln ∣ x - 2 ∣

解答补常数

= 7 ln ∣ x + 1 ∣ - \frac{1}{x + 1} - 3 ln ∣ x - 2 ∣ + C