What is the integral of (e^{3x})/(9+4e) ?

The integral of (e^{3x})/(9+4e) is 1/(3(9+4e))e^{3x}+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})/(9+4e)

\int \frac{e ^{3 x}}{9 + 4 e} d x

\frac{1}{3 ( 9 + 4 e )} e^{3 x} + C

Solution steps

\int \frac{e ^{3 x}}{9 + 4 e} d x

Take the constant out:

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

= \frac{1}{9 + 4 e} \cdot \int e^{3 x} d x

Apply u-substitution

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

= \frac{1}{9 + 4 e} \cdot \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}{9 + 4 e} \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{9 + 4 e} \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 x

= \frac{1}{9 + 4 e} \cdot \frac{1}{3} e^{3 x}

Simplify

\frac{1}{9 + 4 e} \cdot \frac{1}{3} e^{3 x} : \frac{1}{3 ( 9 + 4 e )} e^{3 x}

= \frac{1}{3 ( 9 + 4 e )} e^{3 x}

Add a constant to the solution

= \frac{1}{3 ( 9 + 4 e )} e^{3 x} + C

\int x^{4} cos (x^{5} + 3) d x

\int cot^{5} (\frac{3}{4}) x csc^{9} (\frac{3}{4}) x d x

\int tan^{2} (a x) d x

\int \frac{( 1 - 2 x )}{x - x ^{2}} d x

\int \frac{x ^{2} - 3 x + 5}{x ^{3}} d x

What is the integral of (e^{3x})/(9+4e) ?
The integral of (e^{3x})/(9+4e) is 1/(3(9+4e))e^{3x}+C