What is the integral of (e^{3x})/(3e^{-x)} ?

The integral of (e^{3x})/(3e^{-x)} is 1/12 e^{4x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{3x})/(3e^{-x)}

\int \frac{e ^{3 x}}{3 e ^{- x}} d x

\frac{1}{12} e^{4 x} + C

Solution steps

\int \frac{e ^{3 x}}{3 e ^{- x}} d x

Take the constant out:

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

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

Simplify

\frac{e ^{3 x}}{e ^{- x}} : e^{4 x}

= \frac{1}{3} \cdot \int e^{4 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

= \frac{1}{3} \cdot \int e^{u} \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 x

= \frac{1}{3} \cdot \frac{1}{4} e^{4 x}

Simplify

\frac{1}{3} \cdot \frac{1}{4} e^{4 x} : \frac{1}{12} e^{4 x}

= \frac{1}{12} e^{4 x}

Add a constant to the solution

= \frac{1}{12} e^{4 x} + C

\int \frac{x ^{3}}{9 x ^{2} - 36} d x

\int \frac{x ^{2} + 3}{x ^{2} ( x ^{2} + 9 )} d x

\int x^{2} - 4 x + 10 d x

\int (3 x^{2} + 14 x - 12 + \frac{3}{x ^{2}}) d x

\int \frac{4}{( x - 6 ) ( x - 2 )} d x

What is the integral of (e^{3x})/(3e^{-x)} ?
The integral of (e^{3x})/(3e^{-x)} is 1/12 e^{4x}+C