What is the integral of (e^{2x})/(e^{2x)-5} ?

The integral of (e^{2x})/(e^{2x)-5} is 1/2 ln|e^{2x}-5|+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})/(e^{2x)-5}

\int \frac{e ^{2 x}}{e ^{2 x} - 5} d x

\frac{1}{2} ln e^{2 x} - 5 + C

Solution steps

\int \frac{e ^{2 x}}{e ^{2 x} - 5} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = e^{2 x} - 5

= \frac{1}{2} ln e^{2 x} - 5

Add a constant to the solution

= \frac{1}{2} ln e^{2 x} - 5 + C

\int 6 cos (3 x) sin (3 x) d x

\int (\frac{3 x ^{2}}{2} + 4 x) (4 x + \frac{2}{3}) d x

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

x^{0}

\int ln ∣ t ∣ d t

What is the integral of (e^{2x})/(e^{2x)-5} ?
The integral of (e^{2x})/(e^{2x)-5} is 1/2 ln|e^{2x}-5|+C