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

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

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

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

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

Solution steps

\int e^{3 x + 1} d x

Apply u-substitution

= \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} e^{u}

Substitute back

u = 3 x + 1

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

Add a constant to the solution

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

x \to 27 lim (\frac{x - 27}{x + 9 - 6})

a re a 4 x, x^{2}

\int_{3}^{9} \frac{1}{x ln ( x )} d x

t an g e n t y = 2 x + 1, at x = \frac{1}{2}

\frac{\partial}{\partial x} (\frac{2}{( x - 3 t ) ^{2} + 1})

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