What is the integral of ((8x-3)/(sqrt(12x-4x^2-5))) ?

The integral of ((8x-3)/(sqrt(12x-4x^2-5))) is 1/2 (-4sqrt(-4x^2+12x-5)+9arcsin((2x-3)/2))+C

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integral of ((8x-3)/(sqrt(12x-4x^2-5)))

\int (\frac{8 x - 3}{12 x - 4 x ^{2} - 5}) d x

\frac{1}{2} (- 4 - 4 x^{2} + 12 x - 5 + 9 arcsin (\frac{2 x - 3}{2})) + C

Solution steps

\int \frac{8 x - 3}{12 x - 4 x ^{2} - 5} d x

Apply u-substitution

= \int \frac{u}{2 - u ^{2} + 18 u - 17} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u}{- u ^{2} + 18 u - 17} d u

Complete the square

- u^{2} + 18 u - 17 : - (u - 9)^{2} + 64

= \frac{1}{2} \cdot \int \frac{u}{- ( u - 9 ) ^{2} + 64} d u

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{v + 9}{- v ^{2} + 64} d v

Apply Trigonometric Substitution

= \frac{1}{2} \cdot \int 8 sin (w) + 9 d w

Apply the Sum Rule:

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

= \frac{1}{2} (\int 8 sin (w) d w + \int 9 d w)

\int 8 sin (w) d w = - 8 cos (w)

\int 9 d w = 9 w

= \frac{1}{2} (- 8 cos (w) + 9 w)

Substitute back

= \frac{1}{2} (- 8 cos (arcsin (\frac{1}{8} (8 x - 3 - 9))) + 9 arcsin (\frac{1}{8} (8 x - 3 - 9)))

Simplify

\frac{1}{2} (- 8 cos (arcsin (\frac{1}{8} (8 x - 3 - 9))) + 9 arcsin (\frac{1}{8} (8 x - 3 - 9))) : \frac{1}{2} (- 4 - 4 x^{2} + 12 x - 5 + 9 arcsin (\frac{2 x - 3}{2}))

= \frac{1}{2} (- 4 - 4 x^{2} + 12 x - 5 + 9 arcsin (\frac{2 x - 3}{2}))

Add a constant to the solution

= \frac{1}{2} (- 4 - 4 x^{2} + 12 x - 5 + 9 arcsin (\frac{2 x - 3}{2})) + C

What is the integral of ((8x-3)/(sqrt(12x-4x^2-5))) ?
The integral of ((8x-3)/(sqrt(12x-4x^2-5))) is 1/2 (-4sqrt(-4x^2+12x-5)+9arcsin((2x-3)/2))+C