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

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

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integral of 2xe^{2x^2}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{4} e^{u}

Substitute back

u = 2 x^{2}

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

Simplify

2 \cdot \frac{1}{4} e^{2 x^{2}} : \frac{1}{2} e^{2 x^{2}}

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

Add a constant to the solution

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

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

y^{^{''}} - 10 y^{^{'}} + 25 = 0

\frac{d}{d x} ((\frac{cos ( x )}{1 + sin ( x )})^{5})

\int \frac{x ^{3} + 2 x ^{2} + 3 x - 2}{x - 2} d x

x^{\frac{11}{2}}, 1, x = 1.4

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