What is the integral of (4-32x)/(sqrt(81-16x^2)) ?

The integral of (4-32x)/(sqrt(81-16x^2)) is 2sqrt(-16x^2+81)+arcsin(4/9 x)+C

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integral of (4-32x)/(sqrt(81-16x^2))

\int \frac{4 - 32 x}{81 - 16 x ^{2}} d x

2 - 16 x^{2} + 81 + arcsin (\frac{4}{9} x) + C

Solution steps

\int \frac{4 - 32 x}{81 - 16 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 18 sin (u) + 1 d u

= \int - 18 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 18 sin (u) d u + \int 1 d u

\int 18 sin (u) d u = - 18 cos (u)

\int 1 d u = u

= - (- 18 cos (u)) + u

Substitute back

u = arcsin (\frac{4}{9} x)

= - (- 18 cos (arcsin (\frac{4}{9} x))) + arcsin (\frac{4}{9} x)

Simplify

- (- 18 cos (arcsin (\frac{4}{9} x))) + arcsin (\frac{4}{9} x) : 2 - 16 x^{2} + 81 + arcsin (\frac{4}{9} x)

= 2 - 16 x^{2} + 81 + arcsin (\frac{4}{9} x)

Add a constant to the solution

= 2 - 16 x^{2} + 81 + arcsin (\frac{4}{9} x) + C

x \to 3 lim (\frac{9 + x}{3 - x})

x \to \infty lim (8 x + 5)

f (x) = - x^{2} - 3 x + 4

y^{^{'}} + x^{2} y = 1

\frac{- sin ( x ) 2 x}{3}

What is the integral of (4-32x)/(sqrt(81-16x^2)) ?
The integral of (4-32x)/(sqrt(81-16x^2)) is 2sqrt(-16x^2+81)+arcsin(4/9 x)+C