What is the integral of 1/(sqrt(-x^2-6x-6)) ?

The integral of 1/(sqrt(-x^2-6x-6)) is arcsin(1/(sqrt(3))(x+3))+C

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integral of 1/(sqrt(-x^2-6x-6))

\int \frac{1}{- x ^{2} - 6 x - 6} d x

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

Solution steps

\int \frac{1}{- x ^{2} - 6 x - 6} d x

Complete the square

- x^{2} - 6 x - 6 : - (x + 3)^{2} + 3

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

Apply u-substitution

= \int \frac{1}{- u ^{2} + 3} d u

Apply Trigonometric Substitution

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

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

Simplify

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

Add a constant to the solution

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

What is the integral of 1/(sqrt(-x^2-6x-6)) ?
The integral of 1/(sqrt(-x^2-6x-6)) is arcsin(1/(sqrt(3))(x+3))+C