What is the integral of 1/(sqrt(12x-4x^2-8)) ?

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

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integral of 1/(sqrt(12x-4x^2-8))

\int \frac{1}{12 x - 4 x ^{2} - 8} d x

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

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

\frac{1}{2} \cdot 1 \cdot arcsin (2 (x - \frac{3}{2})) : \frac{1}{2} arcsin (2 x - 3)

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

Add a constant to the solution

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

\frac{d}{d x} (1 - x^{\frac{2}{3}})

\frac{d}{d x} (2 x^{7} - 7 x^{- 4} + 2 x^{2} - 3 x)

x \to \infty lim (\frac{x + 3}{2 x ^{3} - 5})

3^{x} ln (3)

\int_{1}^{2} 4 r^{2} ln (r) d r

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