What is the integral of 1/(sqrt(-x^2-16x)) ?

The integral of 1/(sqrt(-x^2-16x)) is arcsin(1/8 x+1)+C

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integral of 1/(sqrt(-x^2-16x))

\int \frac{1}{- x ^{2} - 16 x} d x

arcsin (\frac{1}{8} x + 1) + C

Solution steps

\int \frac{1}{- x ^{2} - 16 x} d x

Complete the square

- x^{2} - 16 x : - (x + 8)^{2} + 64

= \int \frac{1}{- ( x + 8 ) ^{2} + 64} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 64} d u

= \int \frac{1}{- u ^{2} + 64} 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}{8} (x + 8))

Simplify

1 \cdot arcsin (\frac{1}{8} (x + 8)) : arcsin (\frac{1}{8} x + 1)

= arcsin (\frac{1}{8} x + 1)

Add a constant to the solution

= arcsin (\frac{1}{8} x + 1) + C

\int x 0.04 x^{\frac{3}{4}} + 4 d x

\frac{3}{2} x

\int x \cdot 3 x + 1 d x

\int y^{3} 1 + 9 y^{4} d y

\int coth (5 x) d x

What is the integral of 1/(sqrt(-x^2-16x)) ?
The integral of 1/(sqrt(-x^2-16x)) is arcsin(1/8 x+1)+C