What is the integral of 5/(e^{3x+1)} ?

The integral of 5/(e^{3x+1)} is -5/3 e^{-3x-1}+C

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integral of 5/(e^{3x+1)}

\int \frac{5}{e ^{3 x + 1}} d x

- \frac{5}{3} e^{- 3 x - 1} + C

Solution steps

\int \frac{5}{e ^{3 x + 1}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{e ^{3 x + 1}} d x

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

= 5 \cdot \int \frac{1}{e ^{1}} \cdot \frac{1}{e ^{3 x}} d x

Take the constant out:

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

= 5 \cdot \frac{1}{e ^{1}} \cdot \int \frac{1}{e ^{3 x}} d x

Simplify

= 5 \cdot \frac{1}{e} \cdot \int \frac{1}{e ^{3 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= 5 \cdot \frac{1}{e} \cdot \int e^{- 3 x} d x

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

= 5 \cdot \frac{1}{e} \cdot \int - \frac{1}{3} e^{u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{e} (- \frac{1}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= 5 \cdot \frac{1}{e} (- \frac{1}{3} e^{u})

Substitute back

u = - 3 x

= 5 \cdot \frac{1}{e} (- \frac{1}{3} e^{- 3 x})

Simplify

5 \cdot \frac{1}{e} (- \frac{1}{3} e^{- 3 x}) : - \frac{5}{3} e^{- 3 x - 1}

= - \frac{5}{3} e^{- 3 x - 1}

Add a constant to the solution

= - \frac{5}{3} e^{- 3 x - 1} + C

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What is the integral of 5/(e^{3x+1)} ?
The integral of 5/(e^{3x+1)} is -5/3 e^{-3x-1}+C