What is the integral of (xe^{2x^2}) ?

The integral of (xe^{2x^2}) is 1/4 e^{2x^2}+C

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integral of (xe^{2x^2})

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

\frac{1}{4} e^{2 x^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u}}{4} d u

= \int \frac{e ^{u}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{4} e^{u}

Substitute back

u = 2 x^{2}

= \frac{1}{4} e^{2 x^{2}}

Add a constant to the solution

= \frac{1}{4} e^{2 x^{2}} + C

x \to + 1 + lim (\frac{x ^{2}}{x ^{2} - x})

\frac{d}{d a} (sin (a) + cos (a))

\frac{4}{ln ( x )}

x \to \frac{23}{2} lim (\frac{3}{2})

\int x^{2} ln (7 x) d x

What is the integral of (xe^{2x^2}) ?
The integral of (xe^{2x^2}) is 1/4 e^{2x^2}+C