What is the integral of 1/(e^{2x)-3} ?

The integral of 1/(e^{2x)-3} is 1/6 (ln|e^{2x}-3|-2x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(e^{2x)-3}

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

\frac{1}{6} (ln e^{2 x} - 3 - 2 x) + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{2 u ( u + 3 )} 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 ( u + 3 )} d u

Take the partial fraction of

\frac{1}{u ( u + 3 )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

= \frac{1}{2} \cdot \int \frac{1}{3 u} - \frac{1}{3 ( u + 3 )} 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

= \frac{1}{2} (\int \frac{1}{3 u} d u - \int \frac{1}{3 ( u + 3 )} d u)

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} ln ∣ u + 3 ∣

= \frac{1}{2} (\frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣)

Substitute back

u = e^{2 x} - 3

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

Simplify

\frac{1}{2} (\frac{1}{3} ln e^{2 x} - 3 - \frac{1}{3} ln e^{2 x} - 3 + 3) : \frac{1}{6} (ln e^{2 x} - 3 - 2 x)

= \frac{1}{6} (ln e^{2 x} - 3 - 2 x)

Add a constant to the solution

= \frac{1}{6} (ln e^{2 x} - 3 - 2 x) + C

\int \frac{z ^{2}}{z + 4} d z

\int \frac{x + 2}{x ^{2} - 2 x + 1} d x

\int \frac{x ^{2} + 2 x}{x} d x

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

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

What is the integral of 1/(e^{2x)-3} ?
The integral of 1/(e^{2x)-3} is 1/6 (ln|e^{2x}-3|-2x)+C