What is the integral of (e^x)/(1+81e^{2x)} ?

The integral of (e^x)/(1+81e^{2x)} is 1/9 arctan(9e^x)+C

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integral of (e^x)/(1+81e^{2x)}

\int \frac{e ^{x}}{1 + 81 e ^{2 x}} d x

\frac{1}{9} arctan (9 e^{x}) + C

Solution steps

\int \frac{e ^{x}}{1 + 81 e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{1 + 81 u ^{2}} d u

= \int \frac{1}{1 + 81 u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{9 ( 1 + v ^{2} )} d v

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= \frac{1}{9} arctan (v)

Substitute back

= \frac{1}{9} arctan (9 e^{x})

Add a constant to the solution

= \frac{1}{9} arctan (9 e^{x}) + C

f (x) = 1 + sin (x)

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}

9 x^{2}

f (x) = \frac{x}{2} (1 + (\frac{x}{2})^{2})^{- \frac{1}{2}}

a x^{2} + b x + c

What is the integral of (e^x)/(1+81e^{2x)} ?
The integral of (e^x)/(1+81e^{2x)} is 1/9 arctan(9e^x)+C