What is the integral of (e^x)/(4+9e^{2x)} ?

The integral of (e^x)/(4+9e^{2x)} is 1/6 arctan(3/2 e^x)+C

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integral of (e^x)/(4+9e^{2x)}

\int \frac{e ^{x}}{4 + 9 e ^{2 x}} d x

\frac{1}{6} arctan (\frac{3}{2} e^{x}) + C

Solution steps

\int \frac{e ^{x}}{4 + 9 e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{4 + 9 u ^{2}} d u

= \int \frac{1}{4 + 9 u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{6 ( v ^{2} + 1 )} 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}{v ^{2} + 1} d v

Use the common integral:

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

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

Substitute back

= \frac{1}{6} arctan (\frac{3}{2} e^{x})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{3}{2} e^{x}) + C

- x^{2} + 26 x + 920

y^{^{'}} = \frac{( 1 + x )}{x y ^{6}}

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

f (x) = x^{4} - 5 x^{2} + 2 x - 4, (1, 0)

n = 1 \sum \infty \frac{n}{n ^{3} + 5}

What is the integral of (e^x)/(4+9e^{2x)} ?
The integral of (e^x)/(4+9e^{2x)} is 1/6 arctan(3/2 e^x)+C