What is the integral of (x+1)e^{6x^2+12x} ?

The integral of (x+1)e^{6x^2+12x} is 1/12 e^{6x^2+12x}+C

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integral of (x+1)e^{6x^2+12x}

\int (x + 1) e^{6 x^{2} + 12 x} d x

\frac{1}{12} e^{6 x^{2} + 12 x} + C

Solution steps

\int (x + 1) e^{6 x^{2} + 12 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{12} d u

= \int \frac{e ^{u}}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{12} e^{u}

Substitute back

u = 6 x^{2} + 12 x

= \frac{1}{12} e^{6 x^{2} + 12 x}

Add a constant to the solution

= \frac{1}{12} e^{6 x^{2} + 12 x} + C

\frac{1 - cos ( 2 x )}{1 + sin ( y )} + y^{^{'}} = 0

\frac{d}{d x} (tan (x y))

\frac{\partial}{\partial t} (t - 2 s)

\int 3 z^{3} e^{z} d z

18 x y^{^{'}} - 2 y = - \frac{x ^{2}}{y ^{17}}

What is the integral of (x+1)e^{6x^2+12x} ?
The integral of (x+1)e^{6x^2+12x} is 1/12 e^{6x^2+12x}+C