integral of x^2(7-x^3)^{2/3}

解答

\int x^{2} (7 - x^{3})^{\frac{2}{3}} d x

- \frac{1}{5} (7 - x^{3})^{\frac{5}{3}} + C

求解步骤

\int x^{2} (7 - x^{3})^{\frac{2}{3}} d x

使用换元积分法

= \int - \frac{u ^{\frac{2}{3}}}{3} d u

提出常数:

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

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

使用幂法则

= - \frac{1}{3} \cdot \frac{3}{5} u^{\frac{5}{3}}

u = 7 - x^{3}

代回

= - \frac{1}{3} \cdot \frac{3}{5} (7 - x^{3})^{\frac{5}{3}}

化简

- \frac{1}{3} \cdot \frac{3}{5} (7 - x^{3})^{\frac{5}{3}} : - \frac{1}{5} (7 - x^{3})^{\frac{5}{3}}

= - \frac{1}{5} (7 - x^{3})^{\frac{5}{3}}

解答补常数

= - \frac{1}{5} (7 - x^{3})^{\frac{5}{3}} + C

\frac{d}{d x} (tan (x^{2} - 4 x))

x \to - \infty lim (x^{3} - 5 x^{2})

\int x^{2} \cdot x - 5 d x

d er i v a t i v e y = x^{2} (x^{3} - 1)^{5}

n or ma l y = 2 x + 1, at x = 4