What is the integral of 1/(sqrt(-x^2-18x)) ?

The integral of 1/(sqrt(-x^2-18x)) is arcsin(1/9 x+1)+C

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integral of 1/(sqrt(-x^2-18x))

\int \frac{1}{- x ^{2} - 18 x} d x

arcsin (\frac{1}{9} x + 1) + C

Solution steps

\int \frac{1}{- x ^{2} - 18 x} d x

Complete the square

- x^{2} - 18 x : - (x + 9)^{2} + 81

= \int \frac{1}{- ( x + 9 ) ^{2} + 81} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 81} d u

= \int \frac{1}{- u ^{2} + 81} 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}{9} (x + 9))

Simplify

1 \cdot arcsin (\frac{1}{9} (x + 9)) : arcsin (\frac{1}{9} x + 1)

= arcsin (\frac{1}{9} x + 1)

Add a constant to the solution

= arcsin (\frac{1}{9} x + 1) + C

\int \frac{10}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\int \frac{1}{( x - 5 ) ^{2}} d x

x \to 3 lim (\frac{x - 5}{∣ x - 5 ∣})

(sinh (x))^{^{'}}

n = 1 \sum \infty e^{- n}

What is the integral of 1/(sqrt(-x^2-18x)) ?
The integral of 1/(sqrt(-x^2-18x)) is arcsin(1/9 x+1)+C