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

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

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integral of (x+1)e^{3x^2+6x}

\int (x + 1) e^{3 x^{2} + 6 x} d x

\frac{1}{6} e^{3 x^{2} + 6 x} + C

Solution steps

\int (x + 1) e^{3 x^{2} + 6 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{6} d u

= \int \frac{e ^{u}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{6} e^{u}

Substitute back

u = 3 x^{2} + 6 x

= \frac{1}{6} e^{3 x^{2} + 6 x}

Add a constant to the solution

= \frac{1}{6} e^{3 x^{2} + 6 x} + C

\frac{d}{d x} (A x e^{- 2 x})

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

\frac{\partial}{\partial x} (\frac{x}{x ^{7} - y ^{4}})

\frac{d}{d x} (\frac{7 x ^{2}}{x ^{3}})

x \to 0 lim ((1 + x)^{\frac{7}{x}})

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