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受欢迎的微积分 >

integral of 3t\sqrt[3]{2t^2-11}

解答

\int 3 t 3 2 t^{2} - 11 d t

解答

\frac{9}{16} (2 t^{2} - 11)^{\frac{4}{3}} + C

求解步骤

\int 3 t 3 2 t^{2} - 11 d t

提出常数:

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

= 3 \cdot \int t 3 2 t^{2} - 11 d t

使用换元积分法

= 3 \cdot \int \frac{3 2 u - 11}{2} d u

提出常数:

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

= 3 \cdot \frac{1}{2} \cdot \int 3 2 u - 11 d u

使用换元积分法

= 3 \cdot \frac{1}{2} \cdot \int \frac{3 v}{2} d v

提出常数:

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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int 3 v d v

使用幂法则

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} v^{\frac{4}{3}}

代回

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} (2 t^{2} - 11)^{\frac{4}{3}}

化简

3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} (2 t^{2} - 11)^{\frac{4}{3}} : \frac{9}{16} (2 t^{2} - 11)^{\frac{4}{3}}

= \frac{9}{16} (2 t^{2} - 11)^{\frac{4}{3}}

解答补常数

= \frac{9}{16} (2 t^{2} - 11)^{\frac{4}{3}} + C

作图

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