What is the integral of 1/(sqrt(e^{2x)-6)} ?

The integral of 1/(sqrt(e^{2x)-6)} is 1/(sqrt(6))arctan(sqrt(1/6 (e^{2x)-6)})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(e^{2x)-6)}

\int \frac{1}{e ^{2 x} - 6} d x

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

Solution steps

\int \frac{1}{e ^{2 x} - 6} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 6} d u

= \int \frac{1}{u ^{2} + 6} d u

Apply Integral Substitution

= \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{6} arctan (v)

Substitute back

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

Simplify

\frac{1}{6} arctan (\frac{e ^{2 x} - 6}{6}) : \frac{1}{6} arctan (\frac{1}{6} (e^{2 x} - 6))

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

Add a constant to the solution

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

\int 15 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

\int ln^{6} (x) d x

\int sin (3 x - 5) d x

\frac{d}{d θ} (\frac{5}{sin ( θ ) - 2 cos ( θ )})

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