What is the integral of (x+3)e^{x^2+6x} ?

The integral of (x+3)e^{x^2+6x} is 1/2 e^{x^2+6x}+C

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integral of (x+3)e^{x^2+6x}

\int (x + 3) e^{x^{2} + 6 x} d x

\frac{1}{2} e^{x^{2} + 6 x} + C

Solution steps

\int (x + 3) e^{x^{2} + 6 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} + 6 x

= \frac{1}{2} e^{x^{2} + 6 x}

Add a constant to the solution

= \frac{1}{2} e^{x^{2} + 6 x} + C

x \to 1 + lim (\frac{x - 1}{∣ x - 1 ∣})

2 sec (x) tan (x)

\int cot^{2} (θ) csc (θ) d θ

\frac{s - 1}{( s + 1 ) ( s ^{2} + 1 )}

f (x) = \frac{x - 8}{3 x - 4}, at x = 2

What is the integral of (x+3)e^{x^2+6x} ?
The integral of (x+3)e^{x^2+6x} is 1/2 e^{x^2+6x}+C