integral of (9-w^2)/((w+1)(3w-5)(w+4))

解答

\int \frac{9 - w ^{2}}{( w + 1 ) ( 3 w - 5 ) ( w + 4 )} d w

- \frac{1}{3} ln ∣ w + 1 ∣ + \frac{7}{51} ln ∣ 3 w - 5 ∣ - \frac{7}{51} ln ∣ w + 4 ∣ + C

求解步骤

\int \frac{9 - w ^{2}}{( w + 1 ) ( 3 w - 5 ) ( w + 4 )} d w

将

\frac{9 - w ^{2}}{( w + 1 ) ( 3 w - 5 ) ( w + 4 )}

用部份分式展开:

- \frac{1}{3 ( w + 1 )} + \frac{7}{17 ( 3 w - 5 )} - \frac{7}{51 ( w + 4 )}

= \int - \frac{1}{3 ( w + 1 )} + \frac{7}{17 ( 3 w - 5 )} - \frac{7}{51 ( w + 4 )} d w

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{1}{3 ( w + 1 )} d w + \int \frac{7}{17 ( 3 w - 5 )} d w - \int \frac{7}{51 ( w + 4 )} d w

\int \frac{1}{3 ( w + 1 )} d w = \frac{1}{3} ln ∣ w + 1 ∣

\int \frac{7}{17 ( 3 w - 5 )} d w = \frac{7}{51} ln ∣ 3 w - 5 ∣

\int \frac{7}{51 ( w + 4 )} d w = \frac{7}{51} ln ∣ w + 4 ∣

= - \frac{1}{3} ln ∣ w + 1 ∣ + \frac{7}{51} ln ∣ 3 w - 5 ∣ - \frac{7}{51} ln ∣ w + 4 ∣

解答补常数

= - \frac{1}{3} ln ∣ w + 1 ∣ + \frac{7}{51} ln ∣ 3 w - 5 ∣ - \frac{7}{51} ln ∣ w + 4 ∣ + C

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