What is the integral of (e^x)/(1+36e^{2x)} ?

The integral of (e^x)/(1+36e^{2x)} is 1/6 arctan(6e^x)+C

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integral of (e^x)/(1+36e^{2x)}

\int \frac{e ^{x}}{1 + 36 e ^{2 x}} d x

\frac{1}{6} arctan (6 e^{x}) + C

Solution steps

\int \frac{e ^{x}}{1 + 36 e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{1 + 36 u ^{2}} d u

= \int \frac{1}{1 + 36 u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{6 ( 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}{6} \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}{6} arctan (v)

Substitute back

= \frac{1}{6} arctan (6 e^{x})

Add a constant to the solution

= \frac{1}{6} arctan (6 e^{x}) + C

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\frac{4 x ^{2} + 4 x + 8}{x}

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f (x) = ln (2 x)

\int \frac{x + 1}{x ( x - 2 ) ( x - 3 )} d x

What is the integral of (e^x)/(1+36e^{2x)} ?
The integral of (e^x)/(1+36e^{2x)} is 1/6 arctan(6e^x)+C