What is the integral of 1/(sqrt(12x-9x^2-2)) ?

The integral of 1/(sqrt(12x-9x^2-2)) is 1/3 arcsin(3/(sqrt(2))x-sqrt(2))+C

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integral of 1/(sqrt(12x-9x^2-2))

\int \frac{1}{12 x - 9 x ^{2} - 2} d x

\frac{1}{3} arcsin (\frac{3}{2} x - 2) + C

Solution steps

\int \frac{1}{12 x - 9 x ^{2} - 2} d x

Complete the square

12 x - 9 x^{2} - 2 : - 9 (x - \frac{2}{3})^{2} + 2

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

Apply u-substitution

: \int \frac{1}{- 9 u ^{2} + 2} d u

= \int \frac{1}{- 9 u ^{2} + 2} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{2} (x - \frac{2}{3}))

Simplify

\frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{2} (x - \frac{2}{3})) : \frac{1}{3} arcsin (\frac{3}{2} x - 2)

= \frac{1}{3} arcsin (\frac{3}{2} x - 2)

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{3}{2} x - 2) + C

3 x y

sin (x) ln (2 x)

\frac{\partial}{\partial x} (45 y sin (9 x))

\int_{1}^{3} (y^{4} + \frac{1}{4 y ^{4}}) d y

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = a, y^{^{'}} (0) = b

What is the integral of 1/(sqrt(12x-9x^2-2)) ?
The integral of 1/(sqrt(12x-9x^2-2)) is 1/3 arcsin(3/(sqrt(2))x-sqrt(2))+C