What is the integral of 4/(9e^{2x)+3} ?

The integral of 4/(9e^{2x)+3} is 2/3 (ln(3)+2x-ln|3e^{2x}+1|)+C

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integral of 4/(9e^{2x)+3}

\int \frac{4}{9 e ^{2 x} + 3} d x

\frac{2}{3} (ln (3) + 2 x - ln 3 e^{2 x} + 1) + C

Solution steps

\int \frac{4}{9 e ^{2 x} + 3} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{9 e ^{2 x} + 3} d x

\frac{1}{9 e ^{2 x} + 3} = \frac{1}{3} \frac{1}{3 e ^{2 x} + 1}

= 4 \cdot \int \frac{1}{3} \cdot \frac{1}{3 e ^{2 x} + 1} d x

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{3 e ^{2 x} + 1} d x

Apply u-substitution

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{2 u ( u - 1 )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{u ( u - 1 )} d u

Take the partial fraction of

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

= 4 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int - \frac{1}{u} + \frac{1}{u - 1} 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

= 4 \cdot \frac{1}{3} \cdot \frac{1}{2} (- \int \frac{1}{u} d u + \int \frac{1}{u - 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 4 \cdot \frac{1}{3} \cdot \frac{1}{2} (- ln ∣ u ∣ + ln ∣ u - 1 ∣)

Substitute back

u = 3 e^{2 x} + 1

= 4 \cdot \frac{1}{3} \cdot \frac{1}{2} (- ln 3 e^{2 x} + 1 + ln 3 e^{2 x} + 1 - 1)

Simplify

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

= \frac{2}{3} (ln (3) + 2 x - ln 3 e^{2 x} + 1)

Add a constant to the solution

= \frac{2}{3} (ln (3) + 2 x - ln 3 e^{2 x} + 1) + C

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