What is the integral of ((e^{2x}))/(25+e^{4x)} ?

The integral of ((e^{2x}))/(25+e^{4x)} is 1/10 arctan(1/5 e^{2x})+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}))/(25+e^{4x)}

\int \frac{( e ^{2 x} )}{25 + e ^{4 x}} d x

\frac{1}{10} arctan (\frac{1}{5} e^{2 x}) + C

Solution steps

\int \frac{e ^{2 x}}{25 + e ^{4 x}} d x

Apply u-substitution

: \int \frac{1}{2 ( 25 + u ^{2} )} d u

= \int \frac{1}{2 ( 25 + u ^{2} )} 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}{25 + u ^{2}} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{5 ( 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}{2} \cdot \frac{1}{5} \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}{2} \cdot \frac{1}{5} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{5} arctan (\frac{e ^{2 x}}{5})

Simplify

\frac{1}{2} \cdot \frac{1}{5} arctan (\frac{e ^{2 x}}{5}) : \frac{1}{10} arctan (\frac{1}{5} e^{2 x})

= \frac{1}{10} arctan (\frac{1}{5} e^{2 x})

Add a constant to the solution

= \frac{1}{10} arctan (\frac{1}{5} e^{2 x}) + C

\int e^{x} 7 + e^{x} d x

\int - \frac{1}{262144} cos (8 x) d x

x \to - 3 lim (x^{2} - 8)

\int sec (2 x) tan (2 x) d x

\int x \cdot e^{2 x} d x

What is the integral of ((e^{2x}))/(25+e^{4x)} ?
The integral of ((e^{2x}))/(25+e^{4x)} is 1/10 arctan(1/5 e^{2x})+C