integral of (z^3)/(\sqrt[3]{z^2+1)}

解答

\int \frac{z ^{3}}{3 z ^{2} + 1} d z

\frac{1}{2} (\frac{3}{5} (z^{2} + 1)^{\frac{5}{3}} - \frac{3}{2} (z^{2} + 1)^{\frac{2}{3}}) + C

求解步骤

\int \frac{z ^{3}}{3 z ^{2} + 1} d z

使用换元积分法

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

\int u^{\frac{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

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

= \frac{1}{2} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

u = z^{2} + 1

代回

= \frac{1}{2} (\frac{3}{5} (z^{2} + 1)^{\frac{5}{3}} - \frac{3}{2} (z^{2} + 1)^{\frac{2}{3}})

解答补常数

= \frac{1}{2} (\frac{3}{5} (z^{2} + 1)^{\frac{5}{3}} - \frac{3}{2} (z^{2} + 1)^{\frac{2}{3}}) + C

\int x csc (x^{2}) d x

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 4

\frac{d}{d x} (x - ln (e^{x} cos (x)))

\frac{d}{d x} (\frac{1}{2} x)

d er i v a t i v e f (x) = 4 x^{3}