What is the integral of sqrt(xy-y^2) ?

The integral of sqrt(xy-y^2) is (2(-y^2+yx)^{3/2})/(3y)+C

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integral of sqrt(xy-y^2)

\int x y - y^{2} d x

\frac{2 ( - y ^{2} + y x ) ^{\frac{3}{2}}}{3 y} + C

Solution steps

\int x y - y^{2} d x

Apply u-substitution

: \int \frac{u}{y} d u

= \int \frac{u}{y} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{y} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = - y^{2} + y x

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

Simplify

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

= \frac{2 ( - y ^{2} + y x ) ^{\frac{3}{2}}}{3 y}

Add a constant to the solution

= \frac{2 ( - y ^{2} + y x ) ^{\frac{3}{2}}}{3 y} + C

\frac{d y}{d x} = 0.2 y - 0.001 y^{2}

y = ln (4 t)

y = 2 x^{2} - x^{4}, (1, 1)

\frac{d y}{d x} = - 2 x + 3 y - 5

\frac{\partial}{\partial z} (x + y + z)

What is the integral of sqrt(xy-y^2) ?
The integral of sqrt(xy-y^2) is (2(-y^2+yx)^{3/2})/(3y)+C