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

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

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

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

\frac{1}{3} ln e^{3 x} + 9 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = e^{3 x} + 9

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

Add a constant to the solution

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

\int_{0}^{1} π (6 - \frac{3}{2} x)^{2} d x

\frac{2}{s ^{2} + s}

y^{^{''}} + 9 y = 3 cos (2 x) + 3

\int 9 - x^{2} (- 2 x) d x

\frac{d}{d x} ((x - 2)^{2} (x + 1))

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