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

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

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

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

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

Solution steps

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

Complete the square

- 8 - 4 x^{2} - 12 x : - 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

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