What is the integral of 1/(e^{2x)+3} ?

The integral of 1/(e^{2x)+3} is 1/6 (-ln|e^{2x}+3|+2x)+C

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integral of 1/(e^{2x)+3}

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

\frac{1}{6} (- ln e^{2 x} + 3 + 2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ( u - 3 )} d u

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

Take the partial fraction of

\frac{1}{u ( u - 3 )} : - \frac{1}{3 u} + \frac{1}{3 ( u - 3 )}

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

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u - 3 )} d u = \frac{1}{3} ln ∣ u - 3 ∣

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

Substitute back

u = e^{2 x} + 3

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

Simplify

\frac{1}{2} (- \frac{1}{3} ln e^{2 x} + 3 + \frac{1}{3} ln e^{2 x} + 3 - 3) : \frac{1}{6} (- ln e^{2 x} + 3 + 2 x)

= \frac{1}{6} (- ln e^{2 x} + 3 + 2 x)

Add a constant to the solution

= \frac{1}{6} (- ln e^{2 x} + 3 + 2 x) + C

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What is the integral of 1/(e^{2x)+3} ?
The integral of 1/(e^{2x)+3} is 1/6 (-ln|e^{2x}+3|+2x)+C