integral of (8-128x)/(sqrt(49-64x^2))

解答

\int \frac{8 - 128 x}{49 - 64 x ^{2}} d x

2 - 64 x^{2} + 49 + arcsin (\frac{8}{7} x) + C

求解步骤

\int \frac{8 - 128 x}{49 - 64 x ^{2}} d x

使用三角换元法

= \int - 14 sin (u) + 1 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 14 sin (u) d u + \int 1 d u

\int 14 sin (u) d u = - 14 cos (u)

\int 1 d u = u

= - (- 14 cos (u)) + u

u = arcsin (\frac{8}{7} x)

代回

= - (- 14 cos (arcsin (\frac{8}{7} x))) + arcsin (\frac{8}{7} x)

化简

- (- 14 cos (arcsin (\frac{8}{7} x))) + arcsin (\frac{8}{7} x) : 2 - 64 x^{2} + 49 + arcsin (\frac{8}{7} x)

= 2 - 64 x^{2} + 49 + arcsin (\frac{8}{7} x)

解答补常数

= 2 - 64 x^{2} + 49 + arcsin (\frac{8}{7} x) + C

\frac{\partial}{\partial x} (3 x^{3} + 4 x^{2} y - 3 y^{3})

x \to 0 lim (\frac{3 8 + x - 3 8}{x})

\int \frac{x + 5}{x ^{2} + 10 x + 26} d x

\frac{d y}{d t} = 3 t^{2} cos (t^{3})

d er i v a t i v e x^{- x}