What is the integral of (e^{3x})/(sqrt(4+e^{6x))} ?

The integral of (e^{3x})/(sqrt(4+e^{6x))} is 1/3 ln(1/2 |e^{3x}+sqrt(4+e^{6x)}|)+C

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integral of (e^{3x})/(sqrt(4+e^{6x))}

\int \frac{e ^{3 x}}{4 + e ^{6 x}} d x

\frac{1}{3} ln (\frac{1}{2} e^{3 x} + 4 + e^{6 x}) + C

Solution steps

\int \frac{e ^{3 x}}{4 + e ^{6 x}} d x

Apply u-substitution

: \int \frac{e ^{u}}{3 4 + e ^{2 u}} d u

= \int \frac{e ^{u}}{3 4 + e ^{2 u}} d u

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{4 + v ^{2}} d v

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

Apply Trigonometric Substitution

: \int sec (w) d w

= \frac{1}{3} \cdot \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= \frac{1}{3} ln ∣ tan (w) + sec (w) ∣

Substitute back

= \frac{1}{3} ln tan (arctan (\frac{1}{2} e^{3 x})) + sec (arctan (\frac{1}{2} e^{3 x}))

Simplify

\frac{1}{3} ln tan (arctan (\frac{1}{2} e^{3 x})) + sec (arctan (\frac{1}{2} e^{3 x})) : \frac{1}{3} ln (\frac{1}{2} e^{3 x} + 4 + e^{6 x})

= \frac{1}{3} ln (\frac{1}{2} e^{3 x} + 4 + e^{6 x})

Add a constant to the solution

= \frac{1}{3} ln (\frac{1}{2} e^{3 x} + 4 + e^{6 x}) + C

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What is the integral of (e^{3x})/(sqrt(4+e^{6x))} ?
The integral of (e^{3x})/(sqrt(4+e^{6x))} is 1/3 ln(1/2 |e^{3x}+sqrt(4+e^{6x)}|)+C