integral of (3x^3+6sqrt(x)+1/(x^3))

解答

\int (3 x^{3} + 6 x + \frac{1}{x ^{3}}) d x

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

求解步骤

\int 3 x^{3} + 6 x + \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

= \int 3 x^{3} d x + \int 6 x d x + \int \frac{1}{x ^{3}} d x

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

\int 6 x d x = 4 x^{\frac{3}{2}}

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

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

解答补常数

= \frac{3 x ^{4}}{4} + 4 x^{\frac{3}{2}} - \frac{1}{2 x ^{2}} + C

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