integral of (z^3)/(sqrt(4-z^2))

解答

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

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

求解步骤

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

使用换元积分法

= \int u^{2} - 4 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

= \int u^{2} d u - \int 4 d u

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

\int 4 d u = 4 u

= \frac{u ^{3}}{3} - 4 u

u = 4 - z^{2}

代回

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

化简

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

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

解答补常数

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

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