What is the integral of (2x+8)e^{(x+4)^2} ?

The integral of (2x+8)e^{(x+4)^2} is e^{16}e^{x^2+8x}+C

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integral of (2x+8)e^{(x+4)^2}

\int (2 x + 8) e^{(x + 4)^{2}} d x

e^{16} e^{x^{2} + 8 x} + C

Solution steps

\int (2 x + 8) e^{(x + 4)^{2}} d x

(x + 4)^{2} = x^{2} + 8 x + 16

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

Simplify

e^{x^{2} + 8 x + 16} : e^{x^{2}} e^{8 x} e^{16}

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

Take the constant out:

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

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

Simplify

(2 x + 8) e^{x^{2}} e^{8 x} : e^{x^{2} + 8 x} (2 x + 8)

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

Apply u-substitution

: \int e^{u} d u

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

Use the common integral:

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

= e^{16} e^{u}

Substitute back

u = x^{2} + 8 x

= e^{16} e^{x^{2} + 8 x}

Add a constant to the solution

= e^{16} e^{x^{2} + 8 x} + C

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What is the integral of (2x+8)e^{(x+4)^2} ?
The integral of (2x+8)e^{(x+4)^2} is e^{16}e^{x^2+8x}+C