What is the integral of 1/(sqrt(-9x^2-24x-7)) ?

The integral of 1/(sqrt(-9x^2-24x-7)) is 1/3 arcsin(x+4/3)+C

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integral of 1/(sqrt(-9x^2-24x-7))

\int \frac{1}{- 9 x ^{2} - 24 x - 7} d x

\frac{1}{3} arcsin (x + \frac{4}{3}) + C

Solution steps

\int \frac{1}{- 9 x ^{2} - 24 x - 7} d x

Complete the square

- 9 x^{2} - 24 x - 7 : - 9 (x + \frac{4}{3})^{2} + 9

= \int \frac{1}{- 9 ( x + \frac{4}{3} ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{3 - u ^{2} + 1} d u

= \int \frac{1}{3 - u ^{2} + 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{3} arcsin (u)

Substitute back

u = x + \frac{4}{3}

= \frac{1}{3} arcsin (x + \frac{4}{3})

Add a constant to the solution

= \frac{1}{3} arcsin (x + \frac{4}{3}) + C

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What is the integral of 1/(sqrt(-9x^2-24x-7)) ?
The integral of 1/(sqrt(-9x^2-24x-7)) is 1/3 arcsin(x+4/3)+C