What is the integral of 8xe^{8x^2+3} ?

The integral of 8xe^{8x^2+3} is 1/2 e^{8x^2+3}+C

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integral of 8xe^{8x^2+3}

\int 8 x e^{8 x^{2} + 3} d x

\frac{1}{2} e^{8 x^{2} + 3} + C

Solution steps

\int 8 x e^{8 x^{2} + 3} d x

Take the constant out:

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

= 8 \cdot \int x e^{8 x^{2} + 3} d x

Apply u-substitution

: \int \frac{e ^{u}}{16} d u

= 8 \cdot \int \frac{e ^{u}}{16} d u

Take the constant out:

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

= 8 \cdot \frac{1}{16} \cdot \int e^{u} d u

Use the common integral:

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

= 8 \cdot \frac{1}{16} e^{u}

Substitute back

u = 8 x^{2} + 3

= 8 \cdot \frac{1}{16} e^{8 x^{2} + 3}

Simplify

8 \cdot \frac{1}{16} e^{8 x^{2} + 3} : \frac{1}{2} e^{8 x^{2} + 3}

= \frac{1}{2} e^{8 x^{2} + 3}

Add a constant to the solution

= \frac{1}{2} e^{8 x^{2} + 3} + C

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What is the integral of 8xe^{8x^2+3} ?
The integral of 8xe^{8x^2+3} is 1/2 e^{8x^2+3}+C