integral of 6e^x-x^3(sqrt(x)+1)

解答

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

6 e^{x} - \frac{2}{9} x^{\frac{9}{2}} - \frac{x ^{4}}{4} + C

求解步骤

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

使用换元积分法

= \int \frac{6 u - ln ^{3} ( u ) ( ln ( u ) + 1 )}{u} d u

乘开

\frac{6 u - ln ^{3} ( u ) ( ln ( u ) + 1 )}{u} : 6 - \frac{ln ^{\frac{7}{2}} ( u )}{u} - \frac{ln ^{3} ( u )}{u}

= \int 6 - \frac{ln ^{\frac{7}{2}} ( u )}{u} - \frac{ln ^{3} ( u )}{u} d u

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 6 d u - \int \frac{ln ^{\frac{7}{2}} ( u )}{u} d u - \int \frac{ln ^{3} ( u )}{u} d u

\int 6 d u = 6 u

\int \frac{ln ^{\frac{7}{2}} ( u )}{u} d u = \frac{2}{9} ln^{\frac{9}{2}} (u)

\int \frac{ln ^{3} ( u )}{u} d u = \frac{ln ^{4} ( u )}{4}

= 6 u - \frac{2}{9} ln^{\frac{9}{2}} (u) - \frac{ln ^{4} ( u )}{4}

u = e^{x}

代回

= 6 e^{x} - \frac{2}{9} ln^{\frac{9}{2}} (e^{x}) - \frac{ln ^{4} ( e ^{x} )}{4}

化简

6 e^{x} - \frac{2}{9} ln^{\frac{9}{2}} (e^{x}) - \frac{ln ^{4} ( e ^{x} )}{4} : 6 e^{x} - \frac{2}{9} x^{\frac{9}{2}} - \frac{x ^{4}}{4}

= 6 e^{x} - \frac{2}{9} x^{\frac{9}{2}} - \frac{x ^{4}}{4}

解答补常数

= 6 e^{x} - \frac{2}{9} x^{\frac{9}{2}} - \frac{x ^{4}}{4} + C

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