What is the integral of (e^x)/(1+49e^{2x)} ?

The integral of (e^x)/(1+49e^{2x)} is 1/7 arctan(7e^x)+C

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integral of (e^x)/(1+49e^{2x)}

\int \frac{e ^{x}}{1 + 49 e ^{2 x}} d x

\frac{1}{7} arctan (7 e^{x}) + C

Solution steps

\int \frac{e ^{x}}{1 + 49 e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{1 + 49 u ^{2}} d u

= \int \frac{1}{1 + 49 u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{7 ( 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}{7} \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}{7} arctan (v)

Substitute back

= \frac{1}{7} arctan (7 e^{x})

Add a constant to the solution

= \frac{1}{7} arctan (7 e^{x}) + C

\frac{\partial}{\partial y} (x sin (z))

\frac{\partial}{\partial y} (x^{2} - 2 y^{2})

\frac{d}{d x} (\frac{x}{a})

x \to 0 + lim (ln (arcsin (x)))

x \to \frac{7 π}{2} lim (sec (x))

What is the integral of (e^x)/(1+49e^{2x)} ?
The integral of (e^x)/(1+49e^{2x)} is 1/7 arctan(7e^x)+C