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

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

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integral of (x+1)e^{2x^2+4x}

\int (x + 1) e^{2 x^{2} + 4 x} d x

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

Solution steps

\int (x + 1) e^{2 x^{2} + 4 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{4} d u

= \int \frac{e ^{u}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{4} e^{u}

Substitute back

u = 2 x^{2} + 4 x

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

Add a constant to the solution

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

900 x - \frac{1}{3} x^{3} + 2500

\frac{d}{d x} (arccot (x) + arccot (\frac{1}{x}))

\frac{\partial}{\partial x} (\frac{8}{x})

x \to 0 lim (\frac{x ^{5} + 8 x}{x})

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