What is the integral of 1/(sqrt(3x-x^2)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(3x-x^2))

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 2 \cdot \int \frac{1}{2} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

5 \frac{d x}{d t} + 10 x = 120 sin (60 t)

f (x) = \frac{1}{x - 2}, at x = 6

\int \frac{1}{2 sin ( x ) - cos ( x ) + 5} d x

\frac{d}{d x} ((5)^{x})

\frac{\partial}{\partial x} (x^{n})

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