What is the integral of 1/(sqrt(-x^2-14x-48)) ?

The integral of 1/(sqrt(-x^2-14x-48)) is arcsin(x+7)+C

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integral of 1/(sqrt(-x^2-14x-48))

\int \frac{1}{- x ^{2} - 14 x - 48} d x

arcsin (x + 7) + C

Solution steps

\int \frac{1}{- x ^{2} - 14 x - 48} d x

Complete the square

- x^{2} - 14 x - 48 : - (x + 7)^{2} + 1

= \int \frac{1}{- ( x + 7 ) ^{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 + 7

= arcsin (x + 7)

Add a constant to the solution

= arcsin (x + 7) + C

\int_{0}^{4} 8 x d x

\int \frac{x + 1}{x ^{2} + 2 x} d x

\frac{d y}{d x} = cos (2 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = 6 e^{t}

\frac{d}{d t} (x (t, y)), y = 4 x (t, y)

What is the integral of 1/(sqrt(-x^2-14x-48)) ?
The integral of 1/(sqrt(-x^2-14x-48)) is arcsin(x+7)+C