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

The integral of (6e^{6t})/(3+e^{6t)} is ln|3+e^{6t}|+C

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integral of (6e^{6t})/(3+e^{6t)}

\int \frac{6 e ^{6 t}}{3 + e ^{6 t}} d t

ln 3 + e^{6 t} + C

Solution steps

\int \frac{6 e ^{6 t}}{3 + e ^{6 t}} d t

Take the constant out:

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

= 6 \cdot \int \frac{e ^{6 t}}{3 + e ^{6 t}} d t

Apply u-substitution

: \int \frac{1}{6 u} d u

= 6 \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 + e^{6 t}

= 6 \cdot \frac{1}{6} ln 3 + e^{6 t}

Simplify

6 \cdot \frac{1}{6} ln 3 + e^{6 t} : ln 3 + e^{6 t}

= ln 3 + e^{6 t}

Add a constant to the solution

= ln 3 + e^{6 t} + C

x \to \infty lim (e^{x} + x)

y = a^{3} + cos^{3} (x)

\int \frac{5}{x ^{2} + 8 x + 41} d x

\int 8 sin^{2} (x) cos^{3} (x) d x

y = e^{p} (p + p p)

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