integral of x^3sqrt(x^2+24)

解答

\int x^{3} x^{2} + 24 d x

\frac{1}{5} (x^{2} + 24)^{\frac{5}{2}} - 8 (x^{2} + 24)^{\frac{3}{2}} + C

求解步骤

\int x^{3} x^{2} + 24 d x

使用换元积分法

= \int (u^{2} - 24) u^{2} d u

乘开

(u^{2} - 24) u^{2} : u^{4} - 24 u^{2}

= \int u^{4} - 24 u^{2} 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 u^{4} d u - \int 24 u^{2} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int 24 u^{2} d u = 8 u^{3}

= \frac{u ^{5}}{5} - 8 u^{3}

u = x^{2} + 24

代回

= \frac{( x ^{2} + 24 ) ^{5}}{5} - 8 (x^{2} + 24)^{3}

化简

\frac{( x ^{2} + 24 ) ^{5}}{5} - 8 (x^{2} + 24)^{3} : \frac{1}{5} (x^{2} + 24)^{\frac{5}{2}} - 8 (x^{2} + 24)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 24)^{\frac{5}{2}} - 8 (x^{2} + 24)^{\frac{3}{2}}

解答补常数

= \frac{1}{5} (x^{2} + 24)^{\frac{5}{2}} - 8 (x^{2} + 24)^{\frac{3}{2}} + C

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

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