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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{- 9 u ^{2} + 4} d u

= \int \frac{1}{- 9 u ^{2} + 4} 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 - 1)

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

Add a constant to the solution

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

x \to 0 lim (\frac{x ^{2} + 3 x}{x})

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\frac{\partial}{\partial x} (x y + x z + yz)

\frac{d}{d x} (4 (ln (x) + 1))

\int \frac{7 sin ( x ) + 4 sec ( x )}{tan ( x )} d x

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