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

= \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

\int_{2}^{6} \frac{1}{x ^{2} - 1} d x

\frac{\partial}{\partial y} (e^{y^{2}})

t \cdot \frac{d y}{d t} = t^{3} + 25 t^{3} y

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

y = \frac{2}{3 x ^{3}}