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

The integral of 1/(9+e^{2x)} is 1/18 (-ln|9+e^{2x}|+2x)+C

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

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

\frac{1}{18} (- ln 9 + e^{2 x} + 2 x) + C

Solution steps

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

Apply u-substitution

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

Take the partial fraction of

\frac{1}{u ( u - 9 )} : - \frac{1}{9 u} + \frac{1}{9 ( u - 9 )}

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

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} (- \int \frac{1}{9 u} d u + \int \frac{1}{9 ( u - 9 )} d u)

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u - 9 )} d u = \frac{1}{9} ln ∣ u - 9 ∣

= \frac{1}{2} (- \frac{1}{9} ln ∣ u ∣ + \frac{1}{9} ln ∣ u - 9 ∣)

Substitute back

u = 9 + e^{2 x}

= \frac{1}{2} (- \frac{1}{9} ln 9 + e^{2 x} + \frac{1}{9} ln 9 + e^{2 x} - 9)

Simplify

\frac{1}{2} (- \frac{1}{9} ln 9 + e^{2 x} + \frac{1}{9} ln 9 + e^{2 x} - 9) : \frac{1}{18} (- ln 9 + e^{2 x} + 2 x)

= \frac{1}{18} (- ln 9 + e^{2 x} + 2 x)

Add a constant to the solution

= \frac{1}{18} (- ln 9 + e^{2 x} + 2 x) + C

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What is the integral of 1/(9+e^{2x)} ?
The integral of 1/(9+e^{2x)} is 1/18 (-ln|9+e^{2x}|+2x)+C