integral of w/(sqrt(16-w^2))

解答

\int \frac{w}{16 - w ^{2}} d w

- 16 - w^{2} + C

求解步骤

\int \frac{w}{16 - w ^{2}} d w

使用换元积分法

= \int - \frac{1}{2 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{1}{u} d u

使用幂法则

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

u = 16 - w^{2}

代回

= - \frac{1}{2} \cdot 2 (16 - w^{2})^{\frac{1}{2}}

化简

- \frac{1}{2} \cdot 2 (16 - w^{2})^{\frac{1}{2}} : - 16 - w^{2}

= - 16 - w^{2}

解答补常数

= - 16 - w^{2} + C

\frac{d}{d x} (\frac{x}{x ^{2} + 7})

x \to 0 lim (7 cot^{4} (3 x))

\int \frac{cos ( x )}{x} d x

d er i v a t i v e f (x) = 5 (7 + x)

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