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

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

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

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

2 y - 3 y^{2} + C

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 u} d u

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 y - 3 y^{2}

= \frac{1}{2} \cdot 2 (2 y - 3 y^{2})^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (2 y - 3 y^{2})^{\frac{1}{2}} : 2 y - 3 y^{2}

= 2 y - 3 y^{2}

Add a constant to the solution

= 2 y - 3 y^{2} + C

\frac{d}{d x} (lo g_{2} (1 - 3 x))

(- 6 x^{5} + 12)^{4}

\frac{x ^{2}}{x + 3}

n = 0 \sum \infty 9 + 3 + 1

\frac{d}{d x} ((3 x^{2} - x + 5) (x^{2} + 3 x - 9))

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