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

The integral of (e^x)/(1+9e^{2x)} is 1/3 arctan(3e^x)+C

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integral of (e^x)/(1+9e^{2x)}

\int \frac{e ^{x}}{1 + 9 e ^{2 x}} d x

\frac{1}{3} arctan (3 e^{x}) + C

Solution steps

\int \frac{e ^{x}}{1 + 9 e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{1 + 9 u ^{2}} d u

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

Apply Integral Substitution

= \int \frac{1}{3 ( 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}{3} \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}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (3 e^{x})

Add a constant to the solution

= \frac{1}{3} arctan (3 e^{x}) + C

x \to 3 lim (2 e^{x - 3})

\frac{s}{( s - 1 ) ^{2} + 4}

\frac{d}{d x} (5^{x} + 7^{x})

\frac{d}{d x} (- 262144 cos (2 x))

x \to - 3 lim (h (x))

What is the integral of (e^x)/(1+9e^{2x)} ?
The integral of (e^x)/(1+9e^{2x)} is 1/3 arctan(3e^x)+C