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

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

Popular >

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{e ^{- π s}}{s ^{2} + 4}

\frac{d}{d x} (40 (x^{2} - 2 x))

\int_{2}^{3} \frac{10}{3 - x} d x

\frac{d}{d x} (cos^{- 1} (x))

t^{2}

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