What is the integral of (8-128x)/(sqrt(9-64x^2)) ?

The integral of (8-128x)/(sqrt(9-64x^2)) is 2sqrt(-64x^2+9)+arcsin(8/3 x)+C

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integral of (8-128x)/(sqrt(9-64x^2))

\int \frac{8 - 128 x}{9 - 64 x ^{2}} d x

2 - 64 x^{2} + 9 + arcsin (\frac{8}{3} x) + C

Solution steps

\int \frac{8 - 128 x}{9 - 64 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 6 sin (u) + 1 d u

= \int - 6 sin (u) + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 6 sin (u) d u + \int 1 d u

\int 6 sin (u) d u = - 6 cos (u)

\int 1 d u = u

= - (- 6 cos (u)) + u

Substitute back

u = arcsin (\frac{8}{3} x)

= - (- 6 cos (arcsin (\frac{8}{3} x))) + arcsin (\frac{8}{3} x)

Simplify

- (- 6 cos (arcsin (\frac{8}{3} x))) + arcsin (\frac{8}{3} x) : 2 - 64 x^{2} + 9 + arcsin (\frac{8}{3} x)

= 2 - 64 x^{2} + 9 + arcsin (\frac{8}{3} x)

Add a constant to the solution

= 2 - 64 x^{2} + 9 + arcsin (\frac{8}{3} x) + C

\frac{\partial}{\partial t} (t cos (s))

f (x) = \frac{1}{x ^{2} - 4}

(arccos (x))^{^{'}}

\frac{d ^{4}}{d x ^{4}} (5 e^{x^{2}})

\frac{d}{d x} (x^{3} - 7 x^{2} + 15 x - 10)

What is the integral of (8-128x)/(sqrt(9-64x^2)) ?
The integral of (8-128x)/(sqrt(9-64x^2)) is 2sqrt(-64x^2+9)+arcsin(8/3 x)+C