What is the integral of (e^2x)/((1+3e^2x)) ?

The integral of (e^2x)/((1+3e^2x)) is 1/(9e^2)+1/3 x-1/(9e^2)ln|1+3e^2x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^2x)/((1+3e^2x))

\int \frac{e ^{2} x}{( 1 + 3 e ^{2} x )} d x

\frac{1}{9 e ^{2}} + \frac{1}{3} x - \frac{1}{9 e ^{2}} ln 1 + 3 e^{2} x + C

Solution steps

\int \frac{e ^{2} x}{1 + 3 e ^{2} x} d x

Take the constant out:

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

= e^{2} \cdot \int \frac{x}{1 + 3 e ^{2} x} d x

Apply u-substitution

: \int \frac{u - 1}{9 e ^{4} u} d u

= e^{2} \cdot \int \frac{u - 1}{9 e ^{4} u} d u

Take the constant out:

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

= e^{2} \frac{1}{9 e ^{4}} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= e^{2} \frac{1}{9 e ^{4}} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= e^{2} \frac{1}{9 e ^{4}} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= e^{2} \frac{1}{9 e ^{4}} (u - ln ∣ u ∣)

Substitute back

u = 1 + 3 e^{2} x

= e^{2} \frac{1}{9 e ^{4}} (1 + 3 e^{2} x - ln 1 + 3 e^{2} x)

Simplify

e^{2} \frac{1}{9 e ^{4}} (1 + 3 e^{2} x - ln 1 + 3 e^{2} x) : \frac{1}{9 e ^{2}} + \frac{1}{3} x - \frac{1}{9 e ^{2}} ln 1 + 3 e^{2} x

= \frac{1}{9 e ^{2}} + \frac{1}{3} x - \frac{1}{9 e ^{2}} ln 1 + 3 e^{2} x

Add a constant to the solution

= \frac{1}{9 e ^{2}} + \frac{1}{3} x - \frac{1}{9 e ^{2}} ln 1 + 3 e^{2} x + C

\int \frac{1}{x 49 x ^{2} - 16} d x

\int \frac{4}{x ( 1 + ln ^{2} ( x ) )} d x

\int sin (7 x) e^{3 x} d x

\int \frac{4 x + 6}{5 ( x ^{2} + 9 )} d x

\int \frac{2 x ^{3} + 5 x ^{2} - 4}{x ^{2}} d x

What is the integral of (e^2x)/((1+3e^2x)) ?
The integral of (e^2x)/((1+3e^2x)) is 1/(9e^2)+1/3 x-1/(9e^2)ln|1+3e^2x|+C