What is the integral of (5-4x)/(sqrt(12x-4x^2-8)) ?

The integral of (5-4x)/(sqrt(12x-4x^2-8)) is 2sqrt(-x^2+3x-2)-1/2 arcsin(2x-3)+C

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integral of (5-4x)/(sqrt(12x-4x^2-8))

\int \frac{5 - 4 x}{12 x - 4 x ^{2} - 8} d x

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

Solution steps

\int \frac{5 - 4 x}{12 x - 4 x ^{2} - 8} d x

Complete the square

12 x - 4 x^{2} - 8 : - 4 (x - \frac{3}{2})^{2} + 1

= \int \frac{5 - 4 x}{- 4 ( x - \frac{3}{2} ) ^{2} + 1} d x

Apply u-substitution

: \int \frac{- 4 u - 1}{- 4 u ^{2} + 1} d u

= \int \frac{- 4 u - 1}{- 4 u ^{2} + 1} d u

Expand

\frac{- 4 u - 1}{- 4 u ^{2} + 1} : - \frac{4 u}{- 4 u ^{2} + 1} - \frac{1}{- 4 u ^{2} + 1}

= \int - \frac{4 u}{- 4 u ^{2} + 1} - \frac{1}{- 4 u ^{2} + 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 \frac{4 u}{- 4 u ^{2} + 1} d u - \int \frac{1}{- 4 u ^{2} + 1} d u

\int \frac{4 u}{- 4 u ^{2} + 1} d u = - - 4 u^{2} + 1

\int \frac{1}{- 4 u ^{2} + 1} d u = \frac{1}{2} arcsin (2 u)

= - (- - 4 u^{2} + 1) - \frac{1}{2} arcsin (2 u)

Substitute back

u = x - \frac{3}{2}

= - - - 4 (x - \frac{3}{2})^{2} + 1 - \frac{1}{2} arcsin (2 (x - \frac{3}{2}))

Simplify

- - - 4 (x - \frac{3}{2})^{2} + 1 - \frac{1}{2} arcsin (2 (x - \frac{3}{2})) : 2 - x^{2} + 3 x - 2 - \frac{1}{2} arcsin (2 x - 3)

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

Add a constant to the solution

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

5 (ln (y)) y^{^{'}} - x^{7} y = 0

\int \frac{- 1}{1 + 9 x ^{2}} d x

(- 5, - 5), (0, - 7)

\frac{d}{d x} ((3 x - 2 x^{2})^{3})

x \to \infty lim (1.0 4^{x})

What is the integral of (5-4x)/(sqrt(12x-4x^2-8)) ?
The integral of (5-4x)/(sqrt(12x-4x^2-8)) is 2sqrt(-x^2+3x-2)-1/2 arcsin(2x-3)+C