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

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

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

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

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

Solution steps

\int e^{2 x + 1} d x

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= \int e^{u} \frac{1}{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 = 2 x + 1

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

Add a constant to the solution

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

x y^{^{'}} - y - 3 = 0

x \to 0 lim (\frac{2 tan ( x )}{x})

\int \frac{3}{1 + 3 x} d x

\frac{\partial}{\partial x} (6 x^{2} y)

d y = \frac{x + 6}{x + 1} d x

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