integral of (2-2x)^2

解答

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

- \frac{1}{6} (2 - 2 x)^{3} + C

求解步骤

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

使用换元积分法

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

使用幂法则

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

u = 2 - 2 x

代回

= - \frac{1}{2} \cdot \frac{( 2 - 2 x ) ^{3}}{3}

化简

- \frac{1}{2} \cdot \frac{( 2 - 2 x ) ^{3}}{3} : - \frac{1}{6} (2 - 2 x)^{3}

= - \frac{1}{6} (2 - 2 x)^{3}

解答补常数

= - \frac{1}{6} (2 - 2 x)^{3} + C

d er i v a t i v e \frac{( x ^{2} + 32 x )}{8}

d er i v a t i v e y = 1 - \frac{x ^{2}}{36}

\frac{\partial}{\partial z} (2 y + z)

\int \frac{arcsin ( 5 x )}{1 - 25 x ^{2}} d x

t^{2} y^{^{''}} + 11 t y^{^{'}} + 25 y = 0