integral of sqrt(-4x)

解答

\int - 4 x d x

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

求解步骤

\int - 4 x d x

- 4 x = 4 - x

= \int 4 - x d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int - x d x

4 = 2

= 2 \cdot \int - x d x

使用分布积分法

= 2 (x - x - \int - \frac{x}{2 - x} d x)

\int - \frac{x}{2 - x} d x = \frac{i x ^{\frac{3}{2}}}{3}

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

化简

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

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

解答补常数

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

\int \frac{x}{cos ^{2} ( 2 x )} d x

\int \frac{3}{( x ^{2} + 2 )} d x

\int \frac{x}{a + 2 x} d x

\int 4^{x + \frac{1}{2}} d x

\int y^{2} 4 - y d y