What is the integral of (e^{3x})/(e^{3x)+7} ?

The integral of (e^{3x})/(e^{3x)+7} is 1/3 ln|e^{3x}+7|+C

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integral of (e^{3x})/(e^{3x)+7}

\int \frac{e ^{3 x}}{e ^{3 x} + 7} d x

\frac{1}{3} ln e^{3 x} + 7 + C

Solution steps

\int \frac{e ^{3 x}}{e ^{3 x} + 7} 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} + 7

= \frac{1}{3} ln e^{3 x} + 7

Add a constant to the solution

= \frac{1}{3} ln e^{3 x} + 7 + C

\int x^{2} - e^{2 x} + cos (x) + \frac{3}{1 + x ^{2}} d x

\int (1 - x^{2}) x d x

\int \frac{3 x ^{2} + 2 x - 7}{x ^{2} - 3 x + 2} d x

\int \frac{1}{3 + cos ^{2} ( x )} d x

What is the integral of (e^{3x})/(e^{3x)+7} ?
The integral of (e^{3x})/(e^{3x)+7} is 1/3 ln|e^{3x}+7|+C