What is the integral of (e^x)/(e^{2x)+16} ?

The integral of (e^x)/(e^{2x)+16} is 1/4 arctan((e^x)/4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^x)/(e^{2x)+16}

\int \frac{e ^{x}}{e ^{2 x} + 16} d x

\frac{1}{4} arctan (\frac{e ^{x}}{4}) + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} + 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= \int \frac{1}{u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{4} arctan (\frac{e ^{x}}{4})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{e ^{x}}{4}) + C

\frac{d}{d x} (\frac{x}{∣ x ∣})

\frac{\partial}{\partial x} (π)

- x \frac{d y}{d x} + y = 0

\frac{d ^{2}}{d x ^{2}} (sin (x^{2}))

What is the integral of (e^x)/(e^{2x)+16} ?
The integral of (e^x)/(e^{2x)+16} is 1/4 arctan((e^x)/4)+C