What is the integral of xye^{y^2x} ?

The integral of xye^{y^2x} is 1/2 e^{xy^2}+C

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integral of xye^{y^2x}

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

\frac{1}{2} e^{x y^{2}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{u}}{2 x} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = x y^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{3}{e ^{x} + e ^{- x}}

\frac{d}{d x} (k e^{x})

x sin (\frac{y}{x}) \frac{d y}{d x} = y sin (\frac{y}{x}) + x

\int \frac{1}{a ^{2} + v ^{2}} d v

x \to 4 + lim (\frac{2}{x - 4})

What is the integral of xye^{y^2x} ?
The integral of xye^{y^2x} is 1/2 e^{xy^2}+C