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

\frac{1}{s} \cdot \frac{1}{s ^{2} + \frac{s}{2} + \frac{1}{2}}

\int \frac{1}{x ^{2} + 8 x + 52} d x

x^{2} y^{^{'}} = y^{2} + 3 x y

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

(9, 2.6), (8, 2.2)

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