What is the integral of 4xe^{3x^2-7} ?

The integral of 4xe^{3x^2-7} is 2/3 e^{3x^2-7}+C

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integral of 4xe^{3x^2-7}

\int 4 x e^{3 x^{2} - 7} d x

\frac{2}{3} e^{3 x^{2} - 7} + C

Solution steps

\int 4 x e^{3 x^{2} - 7} d x

Take the constant out:

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

= 4 \cdot \int x e^{3 x^{2} - 7} d x

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{6} e^{u}

Substitute back

u = 3 x^{2} - 7

= 4 \cdot \frac{1}{6} e^{3 x^{2} - 7}

Simplify

4 \cdot \frac{1}{6} e^{3 x^{2} - 7} : \frac{2}{3} e^{3 x^{2} - 7}

= \frac{2}{3} e^{3 x^{2} - 7}

Add a constant to the solution

= \frac{2}{3} e^{3 x^{2} - 7} + C

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What is the integral of 4xe^{3x^2-7} ?
The integral of 4xe^{3x^2-7} is 2/3 e^{3x^2-7}+C