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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

Apply Trigonometric Substitution

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{ln ( x )}{x ( ln ^{2} ( x ) - 1 )} d x

\int \frac{2 x ^{2} - x + 20}{( x - 2 ) ( x ^{2} + 9 )} d x

\int e^{\frac{- t}{2}} d t

\int \frac{( sin ( x ) ) ^{2}}{cos ( x )} d x

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

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