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

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

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

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

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

Solution steps

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

Complete the square

- x^{2} - 6 x : - (x + 3)^{2} + 9

= \int \frac{1}{- ( x + 3 ) ^{2} + 9} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 1 d v

= \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

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

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

Add a constant to the solution

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

x = - y, x = 2 y + 3, y = - 2, y = 1

\frac{d}{d x} (x ln (3))

f (x) = x + 1

\frac{d x}{x ( y - 1 )} + \frac{d y}{y ( x + 2 )} = 0, y (1) = 1

11 x - 4 y x^{2} + 1 y^{^{'}} = 0, y (0) = 7

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