What is the integral of (e^x)/(1+64e^{2x)} ?

The integral of (e^x)/(1+64e^{2x)} is 1/8 arctan(8e^x)+C

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integral of (e^x)/(1+64e^{2x)}

\int \frac{e ^{x}}{1 + 64 e ^{2 x}} d x

\frac{1}{8} arctan (8 e^{x}) + C

Solution steps

\int \frac{e ^{x}}{1 + 64 e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{1 + 64 u ^{2}} d u

= \int \frac{1}{1 + 64 u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{8 ( 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}{8} \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}{8} arctan (v)

Substitute back

= \frac{1}{8} arctan (8 e^{x})

Add a constant to the solution

= \frac{1}{8} arctan (8 e^{x}) + C

e^{3 x} \cdot ln (4 x)

x \to 2 lim (5 - 3 x)

\frac{d}{d x} (cos (2 x - 1) \cdot tan (1 - 2 x))

\int 3 x (2 - 3 x^{2}) d x

What is the integral of (e^x)/(1+64e^{2x)} ?
The integral of (e^x)/(1+64e^{2x)} is 1/8 arctan(8e^x)+C