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

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

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

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

\frac{1}{2} e^{x^{2} + 4 x} + C

Solution steps

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

Apply u-substitution

= \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} + 4 x

= \frac{1}{2} e^{x^{2} + 4 x}

Add a constant to the solution

= \frac{1}{2} e^{x^{2} + 4 x} + C

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