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

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

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integral of (e^x)/(1+4e^{2x)}

\int \frac{e ^{x}}{1 + 4 e ^{2 x}} d x

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

Solution steps

\int \frac{e ^{x}}{1 + 4 e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{1 + 4 u ^{2}} d u

= \int \frac{1}{1 + 4 u ^{2}} d u

Apply Integral Substitution

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

Substitute back

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

Add a constant to the solution

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

d x - \frac{1}{y ^{2} - 6 y + 13} d y = 0

cos (2 x)

\frac{d}{d x} (sec (x) sin (x))

\int \frac{ln ( x )}{x ^{2}} d x

y = \frac{3}{x + 1}, (2, 1)

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