What is the integral of 1/(sqrt(-x^2-10x-24)) ?

The integral of 1/(sqrt(-x^2-10x-24)) is arcsin(x+5)+C

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integral of 1/(sqrt(-x^2-10x-24))

\int \frac{1}{- x ^{2} - 10 x - 24} d x

arcsin (x + 5) + C

Solution steps

\int \frac{1}{- x ^{2} - 10 x - 24} d x

Complete the square

- x^{2} - 10 x - 24 : - (x + 5)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = arcsin (u)

= arcsin (u)

Substitute back

u = x + 5

= arcsin (x + 5)

Add a constant to the solution

= arcsin (x + 5) + C

x \to 3 lim (1 - x - x^{2})

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{9 x ^{\frac{8}{9}}})

\int_{1}^{2} \frac{x - 1}{x} d x

\int e^{x} \cdot 2 x d x

\int \frac{t ^{2}}{1 + t ^{6}} d t

What is the integral of 1/(sqrt(-x^2-10x-24)) ?
The integral of 1/(sqrt(-x^2-10x-24)) is arcsin(x+5)+C