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

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

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integral of (e^{3x})/(9+e^{2x)}

\int \frac{e ^{3 x}}{9 + e ^{2 x}} d x

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

Solution steps

\int \frac{e ^{3 x}}{9 + e ^{2 x}} d x

Apply u-substitution

: \int \frac{u - 9}{2 u} d u

= \int \frac{u - 9}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u - 9}{u} d u

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} + 9} d v

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

\frac{v ^{2}}{v ^{2} + 9} = - \frac{9}{v ^{2} + 9} + 1

= \frac{1}{2} \cdot 2 \cdot \int - \frac{9}{v ^{2} + 9} + 1 d v

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int 3 (- \frac{1}{w ^{2} + 1} + 1) d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot 2 \cdot 3 \cdot \int - \frac{1}{w ^{2} + 1} + 1 d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot 2 \cdot 3 (- \int \frac{1}{w ^{2} + 1} d w + \int 1 d w)

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

\int 1 d w = w

= \frac{1}{2} \cdot 2 \cdot 3 (- arctan (w) + w)

Substitute back

= \frac{1}{2} \cdot 2 \cdot 3 (- arctan (\frac{9 + e ^{2 x} - 9}{3}) + \frac{9 + e ^{2 x} - 9}{3})

Simplify

\frac{1}{2} \cdot 2 \cdot 3 (- arctan (\frac{9 + e ^{2 x} - 9}{3}) + \frac{9 + e ^{2 x} - 9}{3}) : - 3 arctan (\frac{1}{3} e^{x}) + e^{x}

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

Add a constant to the solution

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

\int_{0}^{3} (\frac{6}{1 + x ^{2}}) d x

x \to 0 lim ((3 x - 4)^{4})

2 e^{\frac{s i n ( w )}{3}} + 1

\frac{d}{d x} (x^{2} 4 x - 3)

x \to 4 - lim (x^{2} + 2)

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