What is the integral of (e^w)/(36+e^{2w)} ?

The integral of (e^w)/(36+e^{2w)} is 1/6 arctan((e^w)/6)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (e^w)/(36+e^{2w)}

\int \frac{e ^{w}}{36 + e ^{2 w}} d w

\frac{1}{6} arctan (\frac{e ^{w}}{6}) + C

Solution steps

\int \frac{e ^{w}}{36 + e ^{2 w}} d w

Apply u-substitution

= \int \frac{1}{36 + u ^{2}} 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 ^{w}}{6})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{e ^{w}}{6}) + C

What is the integral of (e^w)/(36+e^{2w)} ?
The integral of (e^w)/(36+e^{2w)} is 1/6 arctan((e^w)/6)+C