integral of (z^2)/(\sqrt[3]{4+z^3)}

解答

\int \frac{z ^{2}}{3 4 + z ^{3}} d z

\frac{1}{2} (4 + z^{3})^{\frac{2}{3}} + C

求解步骤

\int \frac{z ^{2}}{3 4 + z ^{3}} d z

使用换元积分法

= \int \frac{1}{3 3 u} 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 \frac{1}{3 u} d u

使用幂法则

= \frac{1}{3} \cdot \frac{3}{2} u^{\frac{2}{3}}

u = 4 + z^{3}

代回

= \frac{1}{3} \cdot \frac{3}{2} (4 + z^{3})^{\frac{2}{3}}

化简

\frac{1}{3} \cdot \frac{3}{2} (4 + z^{3})^{\frac{2}{3}} : \frac{1}{2} (4 + z^{3})^{\frac{2}{3}}

= \frac{1}{2} (4 + z^{3})^{\frac{2}{3}}

解答补常数

= \frac{1}{2} (4 + z^{3})^{\frac{2}{3}} + C

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

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