What is the integral of (2x-3x^2)^{-0.5} ?

The integral of (2x-3x^2)^{-0.5} is 1/(sqrt(3))arcsin(3x-1)+C

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integral of (2x-3x^2)^{-0.5}

\int (2 x - 3 x^{2})^{- 0.5} d x

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

Solution steps

\int (2 x - 3 x^{2})^{- 0.5} d x

Complete the square

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

= \int (- 3 (x - \frac{1}{3})^{2} + \frac{1}{3})^{- 0.5} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= 3 \cdot \int \frac{1}{3} d v

Take the constant out:

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

= 3 \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 3 \frac{1}{3} \cdot 1 \cdot v

Substitute back

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

Simplify

3 \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_{0}^{\infty} y^{2} e^{- y} d y

\int \frac{x ^{4}}{x ^{3} + x ^{2} - x - 1} d x

\int \frac{s}{2 s + 3} d s

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 2

\int x (3 + 5 x) d x

What is the integral of (2x-3x^2)^{-0.5} ?
The integral of (2x-3x^2)^{-0.5} is 1/(sqrt(3))arcsin(3x-1)+C