What is the integral of (e^{2x})/(1+6e^{2x)+9e^{2x}} ?

The integral of (e^{2x})/(1+6e^{2x)+9e^{2x}} is 1/30 ln|1+15e^{2x}|+C

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integral of (e^{2x})/(1+6e^{2x)+9e^{2x}}

\int \frac{e ^{2 x}}{1 + 6 e ^{2 x} + 9 e ^{2 x}} d x

\frac{1}{30} ln 1 + 15 e^{2 x} + C

Solution steps

\int \frac{e ^{2 x}}{1 + 6 e ^{2 x} + 9 e ^{2 x}} d x

Simplify

= \int \frac{e ^{2 x}}{1 + 15 e ^{2 x}} d x

Apply u-substitution

= \int \frac{1}{30 u} d u

Take the constant out:

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

= \frac{1}{30} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{30} ln ∣ u ∣

Substitute back

u = 1 + 15 e^{2 x}

= \frac{1}{30} ln 1 + 15 e^{2 x}

Add a constant to the solution

= \frac{1}{30} ln 1 + 15 e^{2 x} + C

What is the integral of (e^{2x})/(1+6e^{2x)+9e^{2x}} ?
The integral of (e^{2x})/(1+6e^{2x)+9e^{2x}} is 1/30 ln|1+15e^{2x}|+C