What is the integral of e^{3x+8} ?

The integral of e^{3x+8} is 1/3 e^{3x+8}+C

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integral of e^{3x+8}

\int e^{3 x + 8} d x

\frac{1}{3} e^{3 x + 8} + C

Solution steps

\int e^{3 x + 8} 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 + 8

= \frac{1}{3} e^{3 x + 8}

Add a constant to the solution

= \frac{1}{3} e^{3 x + 8} + C

\frac{d}{d x} (\frac{cot ( x )}{1 + csc ( x )})

\frac{\partial}{\partial y} (x^{2} + 2 y + 2 z)

c^{^{'}}

\frac{d ^{4}}{d x ^{4}} (3 x^{4})

\frac{4}{x + 1}