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

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

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

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

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

Solution steps

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

Complete the square

8 + 6 x - 9 x^{2} : - 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

= \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

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