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

解答

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

\frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln x^{2} + 3 - \frac{1}{3} arctan (\frac{x}{3})) + C

求解步骤

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

将

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

用部份分式展开:

\frac{1}{4 ( x - 1 )} + \frac{- x - 1}{4 ( x ^{2} + 3 )}

= \int \frac{1}{4 ( x - 1 )} + \frac{- x - 1}{4 ( x ^{2} + 3 )} 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{1}{4 ( x - 1 )} d x + \int \frac{- x - 1}{4 ( x ^{2} + 3 )} d x

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{- x - 1}{4 ( x ^{2} + 3 )} d x = \frac{1}{4} (- \frac{1}{2} ln x^{2} + 3 - \frac{1}{3} arctan (\frac{x}{3}))

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln x^{2} + 3 - \frac{1}{3} arctan (\frac{x}{3}))

解答补常数

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln x^{2} + 3 - \frac{1}{3} arctan (\frac{x}{3})) + C

\int y e^{x y} cos (2 x) d x

\int 3 t^{2} d t

\frac{d}{d z} (\frac{z}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

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x y^{^{'}} = y + 2 x^{3} sin^{2} (\frac{y}{x})