What is the integral of (2-8x)/(sqrt(9-4x^2)) ?

The integral of (2-8x)/(sqrt(9-4x^2)) is 2sqrt(-4x^2+9)+arcsin(2/3 x)+C

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integral of (2-8x)/(sqrt(9-4x^2))

\int \frac{2 - 8 x}{9 - 4 x ^{2}} d x

2 - 4 x^{2} + 9 + arcsin (\frac{2}{3} x) + C

Solution steps

\int \frac{2 - 8 x}{9 - 4 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{2}{3} x)

= - (- 6 cos (arcsin (\frac{2}{3} x))) + arcsin (\frac{2}{3} x)

Simplify

- (- 6 cos (arcsin (\frac{2}{3} x))) + arcsin (\frac{2}{3} x) : 2 - 4 x^{2} + 9 + arcsin (\frac{2}{3} x)

= 2 - 4 x^{2} + 9 + arcsin (\frac{2}{3} x)

Add a constant to the solution

= 2 - 4 x^{2} + 9 + arcsin (\frac{2}{3} x) + C

y = 2 x^{2} - 9 x + 7

\frac{d}{d x} (\frac{1}{2 cos ^{2} ( x )})

\int (x x + 2) d x

\frac{d}{d x} (2 x - 0.1 x^{2})

y = 9 + 4 x^{2} - 2 x^{3}, (1, 11)

What is the integral of (2-8x)/(sqrt(9-4x^2)) ?
The integral of (2-8x)/(sqrt(9-4x^2)) is 2sqrt(-4x^2+9)+arcsin(2/3 x)+C