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

The integral of (x+4)*e^{x^2+8x} is 1/2 e^{x^2+8x}+C

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = x^{2} + 8 x

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

Add a constant to the solution

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

x^{3} - 8

x (7^{- 3 x})

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

x \to 2 lim (16)

n = 1 \sum \infty 5 (4)^{n - 1}

What is the integral of (x+4)*e^{x^2+8x} ?
The integral of (x+4)*e^{x^2+8x} is 1/2 e^{x^2+8x}+C