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

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

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

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

arcsin (\frac{1}{4} x + 1) + C

Solution steps

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

Complete the square

- x^{2} - 8 x : - (x + 4)^{2} + 16

= \int \frac{1}{- ( x + 4 ) ^{2} + 16} d x

Apply u-substitution

= \int \frac{1}{- u ^{2} + 16} 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}{4} (x + 4))

Simplify

1 \cdot arcsin (\frac{1}{4} (x + 4)) : arcsin (\frac{1}{4} x + 1)

= arcsin (\frac{1}{4} x + 1)

Add a constant to the solution

= arcsin (\frac{1}{4} x + 1) + C

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