What is the integral of xye^{xy^2} ?

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

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integral of xye^{xy^2}

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

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

Solution steps

\int x y e^{x y^{2}} 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{d}{d x} (\frac{6 x}{x ^{2} - 4})

\int_{0}^{N} e^{- s t} e^{- 4 t} d t

x \to + 0 + lim (\frac{5 x}{sin ( x )})

\frac{1}{x} \cdot 3 e^{x}

\int \frac{1}{9 x ^{2} - 4} d x

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