integral from 0 to sqrt(7 of)(x^3)/(\sqrt[3]{1+x^2)}

解答

\int_{0}^{7} \frac{x ^{3}}{3 1 + x ^{2}} d x

\frac{141}{20}

十进制

7.05

求解步骤

\int_{0}^{7} \frac{x ^{3}}{3 1 + x ^{2}} d x

使用换元积分法

= \int_{1}^{8} \frac{u - 1}{2 3 u} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{1}^{8} \frac{u - 1}{3 u} d u

乘开

\frac{u - 1}{3 u} : u^{\frac{2}{3}} - \frac{1}{3 u}

= \frac{1}{2} \cdot \int_{1}^{8} u^{\frac{2}{3}} - \frac{1}{3 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

= \frac{1}{2} (\int_{1}^{8} u^{\frac{2}{3}} d u - \int_{1}^{8} \frac{1}{3 u} d u)

\int_{1}^{8} u^{\frac{2}{3}} d u = \frac{93}{5}

\int_{1}^{8} \frac{1}{3 u} d u = \frac{9}{2}

= \frac{1}{2} (\frac{93}{5} - \frac{9}{2})

化简

\frac{1}{2} (\frac{93}{5} - \frac{9}{2}) : \frac{141}{20}

= \frac{141}{20}

x \to 3 lim (\frac{( - 4 ) ( 0 )}{4})

d er i v a t i v e \frac{x ^{7} + 2}{x ^{7}}

t an g e n t 11 - x^{2} (3.2)

\int tan^{4} (6 x) sec^{2} (6 x) d x

\frac{\partial}{\partial y} (y^{2} - 9)