integral of 2\sqrt[5]{x^4}-7x^3+10e^x-1

解答

\int 2 5 x^{4} - 7 x^{3} + 10 e^{x} - 1 d x

\frac{10}{9} x^{\frac{9}{5}} - \frac{7 x ^{4}}{4} + 10 e^{x} - x + C

求解步骤

\int 2 5 x^{4} - 7 x^{3} + 10 e^{x} - 1 d x

5 x^{4} = x^{\frac{4}{5}},

假定

x \geq 0

= \int 2 x^{\frac{4}{5}} - 7 x^{3} + 10 e^{x} - 1 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 2 x^{\frac{4}{5}} d x - \int 7 x^{3} d x + \int 10 e^{x} d x - \int 1 d x

\int 2 x^{\frac{4}{5}} d x = \frac{10}{9} x^{\frac{9}{5}}

\int 7 x^{3} d x = \frac{7 x ^{4}}{4}

\int 10 e^{x} d x = 10 e^{x}

\int 1 d x = x

= \frac{10}{9} x^{\frac{9}{5}} - \frac{7 x ^{4}}{4} + 10 e^{x} - x

解答补常数

= \frac{10}{9} x^{\frac{9}{5}} - \frac{7 x ^{4}}{4} + 10 e^{x} - x + C

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