What is the integral of 24y(x-yx-(x^2)/2) ?

The integral of 24y(x-yx-(x^2)/2) is 24(-(x^2y^2)/4+(xy^2)/2-(xy^3)/3)+C

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integral of 24y(x-yx-(x^2)/2)

\int 24 y (x - y x - \frac{x ^{2}}{2}) d y

24 (- \frac{x ^{2} y ^{2}}{4} + \frac{x y ^{2}}{2} - \frac{x y ^{3}}{3}) + C

Solution steps

\int 24 y (x - y x - \frac{x ^{2}}{2}) d y

Take the constant out:

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

= 24 \cdot \int (- \frac{x ^{2}}{2} + x - x y) y d y

Expand

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

= 24 \cdot \int - \frac{x ^{2} y}{2} + x y - x y^{2} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 24 (- \int \frac{x ^{2} y}{2} d y + \int x y d y - \int x y^{2} d y)

\int \frac{x ^{2} y}{2} d y = \frac{x ^{2} y ^{2}}{4}

\int x y d y = \frac{x y ^{2}}{2}

\int x y^{2} d y = \frac{x y ^{3}}{3}

= 24 (- \frac{x ^{2} y ^{2}}{4} + \frac{x y ^{2}}{2} - \frac{x y ^{3}}{3})

Add a constant to the solution

= 24 (- \frac{x ^{2} y ^{2}}{4} + \frac{x y ^{2}}{2} - \frac{x y ^{3}}{3}) + C

\int x (x + 3) d x

2 x e^{x}

\int 2 t e^{- t^{2}} d t

\int_{0}^{1} x 1 - x^{2} d x

\int \frac{2 u - 1}{- u + 2} d u

What is the integral of 24y(x-yx-(x^2)/2) ?
The integral of 24y(x-yx-(x^2)/2) is 24(-(x^2y^2)/4+(xy^2)/2-(xy^3)/3)+C