What is the integral of ((x+1)e^{8x^2+16x}) ?

The integral of ((x+1)e^{8x^2+16x}) is 1/16 e^{8x^2+16x}+C

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integral of ((x+1)e^{8x^2+16x})

\int ((x + 1) e^{8 x^{2} + 16 x}) d x

\frac{1}{16} e^{8 x^{2} + 16 x} + C

Solution steps

\int (x + 1) e^{8 x^{2} + 16 x} d x

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = 8 x^{2} + 16 x

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

Add a constant to the solution

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

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What is the integral of ((x+1)e^{8x^2+16x}) ?
The integral of ((x+1)e^{8x^2+16x}) is 1/16 e^{8x^2+16x}+C