What is the integral of 1/(sqrt(-9x^2+6x+8)) ?

The integral of 1/(sqrt(-9x^2+6x+8)) is 1/3 arcsin(x-1/3)+C

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integral of 1/(sqrt(-9x^2+6x+8))

\int \frac{1}{- 9 x ^{2} + 6 x + 8} d x

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

Solution steps

\int \frac{1}{- 9 x ^{2} + 6 x + 8} d x

Complete the square

- 9 x^{2} + 6 x + 8 : - 9 (x - \frac{1}{3})^{2} + 9

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

Apply u-substitution

= \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{1}{3}

= \frac{1}{3} arcsin (x - \frac{1}{3})

Add a constant to the solution

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

What is the integral of 1/(sqrt(-9x^2+6x+8)) ?
The integral of 1/(sqrt(-9x^2+6x+8)) is 1/3 arcsin(x-1/3)+C