What is the integral of 2xe^{3x^2} ?

The integral of 2xe^{3x^2} is 1/3 e^{3x^2}+C

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integral of 2xe^{3x^2}

\int 2 x e^{3 x^{2}} d x

\frac{1}{3} e^{3 x^{2}} + C

Solution steps

\int 2 x e^{3 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^{3 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{6} d u

= 2 \cdot \int \frac{e ^{u}}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{6} e^{u}

Substitute back

u = 3 x^{2}

= 2 \cdot \frac{1}{6} e^{3 x^{2}}

Simplify

2 \cdot \frac{1}{6} e^{3 x^{2}} : \frac{1}{3} e^{3 x^{2}}

= \frac{1}{3} e^{3 x^{2}}

Add a constant to the solution

= \frac{1}{3} e^{3 x^{2}} + C

x \to + (- 1) lim (x - 1)

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\frac{\partial}{\partial x} (- x^{3} + 6 x^{2} y^{3} + 5 y^{2})

\int \frac{2 x + 3}{( 2 x ^{2} + 6 x ) ^{4}} d x

What is the integral of 2xe^{3x^2} ?
The integral of 2xe^{3x^2} is 1/3 e^{3x^2}+C