integral of (x^4)/((x^2-1)(x-2))

解答

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

\frac{x ^{2}}{2} + 2 x + \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣ + C

求解步骤

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

将

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

用部份分式展开:

x + 2 + \frac{1}{6 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{16}{3 ( x - 2 )}

= \int x + 2 + \frac{1}{6 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{16}{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 x d x + \int 2 d x + \int \frac{1}{6 ( x + 1 )} d x - \int \frac{1}{2 ( x - 1 )} d x + \int \frac{16}{3 ( x - 2 )} d x

\int x d x = \frac{x ^{2}}{2}

\int 2 d x = 2 x

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

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

\int \frac{16}{3 ( x - 2 )} d x = \frac{16}{3} ln ∣ x - 2 ∣

= \frac{x ^{2}}{2} + 2 x + \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣

解答补常数

= \frac{x ^{2}}{2} + 2 x + \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣ + C

\int \frac{sinh ( e ^{- \frac{x}{2}} )}{e ^{\frac{x}{2}}} d x

\int r ln (\frac{2 r}{g}) d r

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\int (cos (x) + x) d x

\int \frac{1 + x ^{2}}{x ^{4}} d x