integral of 4x+sin(3x)

解答

\int 4 x + sin (3 x) d x

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

求解步骤

\int 4 x + sin (3 x) d x

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 4 x d x + \int sin (3 x) d x

\int 4 x d x = 2 x^{2}

\int sin (3 x) d x = - \frac{1}{3} cos (3 x)

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

解答补常数

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

s l o p e in t erce pt (6.4) (2.1)

\frac{\partial}{\partial x} (- 2 e^{- 3 x^{2} - y^{2}} y)

d er i v a t i v e y = 1.00059 e^{(2.39736)} - 1.0005

\frac{d}{d x} (\frac{1}{2 x + 7})

d er i v a t i v e f (x) = x \cdot e^{2 x - 4}