What is the integral of 3/(sqrt(6x-x^2)) ?

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

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integral of 3/(sqrt(6x-x^2))

\int \frac{3}{6 x - x ^{2}} d x

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

Solution steps

\int \frac{3}{6 x - x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{6 x - x ^{2}} d x

Complete the square

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

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

Apply u-substitution

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

= 3 \cdot \int \frac{1}{- u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int 1 d v

= 3 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 3 \cdot 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 5 x + 6, at a = 2

\frac{d}{d x} (ln (x^{2} - 14 x))

2 x^{2} - 80 x + y^{2} + 49 = 0 (0)

\int_{1}^{2} 4 - x^{2} d x

\int x^{2} \cdot sin (4 x) d x

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