integral of 6t^2*\sqrt[3]{t}

解答

\int 6 t^{2} \cdot 3 t d t

6 (\frac{1}{3} t^{3} 3 t - \frac{1}{30} t^{\frac{10}{3}}) + C

求解步骤

\int 6 t^{2} 3 t d t

提出常数:

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

= 6 \cdot \int t^{2} 3 t d t

使用分布积分法

= 6 (\frac{1}{3} t^{3} 3 t - \int \frac{1}{9} t^{\frac{7}{3}} d t)

\int \frac{1}{9} t^{\frac{7}{3}} d t = \frac{1}{30} t^{\frac{10}{3}}

= 6 (\frac{1}{3} t^{3} 3 t - \frac{1}{30} t^{\frac{10}{3}})

解答补常数

= 6 (\frac{1}{3} t^{3} 3 t - \frac{1}{30} t^{\frac{10}{3}}) + C

\int 5 x tan (x^{2}) d x

\int 4 x (ln (x^{9}))^{2} d x

\int (cos (2 x) - sin (2 x)) d x

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

\int (sin (4 x)) e^{x} d x