integral of ((x^3-4x-1))/(2x^3)

解答

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

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

求解步骤

\int \frac{x ^{3} - 4 x - 1}{2 x ^{3}} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{x ^{3} - 4 x - 1}{x ^{3}} d x

乘开

\frac{x ^{3} - 4 x - 1}{x ^{3}} : 1 - \frac{4}{x ^{2}} - \frac{1}{x ^{3}}

= \frac{1}{2} \cdot \int 1 - \frac{4}{x ^{2}} - \frac{1}{x ^{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

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

\int 1 d x = x

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

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

化简

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

解答补常数

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

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