What is the integral of (y^2e^{xy}) ?

The integral of (y^2e^{xy}) is ye^{yx}+C

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integral of (y^2e^{xy})

\int (y^{2} e^{x y}) d x

y e^{y x} + C

Solution steps

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

Take the constant out:

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

= y^{2} \cdot \int e^{y x} d x

Apply u-substitution

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

= y^{2} \cdot \int \frac{e ^{u}}{y} d u

Take the constant out:

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

= y^{2} \frac{1}{y} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= y^{2} \frac{1}{y} e^{u}

Substitute back

u = y x

= y^{2} \frac{1}{y} e^{y x}

Simplify

y^{2} \frac{1}{y} e^{y x} : y e^{y x}

= y e^{y x}

Add a constant to the solution

= y e^{y x} + C

\frac{\partial}{\partial x} (\frac{x - 2 y}{x + 2 y})

\frac{\partial}{\partial x} (ln (x + 2 c t))

f (x) = 186.5

\int \frac{2}{x ln ( 6 x )} d x

\int \frac{1}{x 6 x + 49} d x