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

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

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integral of (3e^{3t})/(3+e^{3t)}

\int \frac{3 e ^{3 t}}{3 + e ^{3 t}} d t

ln 3 + e^{3 t} + C

Solution steps

\int \frac{3 e ^{3 t}}{3 + e ^{3 t}} d t

Take the constant out:

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

= 3 \cdot \int \frac{e ^{3 t}}{3 + e ^{3 t}} d t

Apply u-substitution

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

= 3 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 + e^{3 t}

= 3 \cdot \frac{1}{3} ln 3 + e^{3 t}

Simplify

3 \cdot \frac{1}{3} ln 3 + e^{3 t} : ln 3 + e^{3 t}

= ln 3 + e^{3 t}

Add a constant to the solution

= ln 3 + e^{3 t} + C

y = ((25 - x^{2})^{\frac{1}{2}}), y = \frac{4}{3} x, x = 5

(\frac{1 + x ^{2}}{1 - x ^{2}})^{17}

\frac{d y}{d x} = sin (17 x)

x \to 1 lim (\frac{4 - x}{x - 1})

\int (x + 1) \cdot e^{4 x + 1} d x

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