What is the integral of (6e^{6t})/(9+e^{6t)} ?

The integral of (6e^{6t})/(9+e^{6t)} is ln|9+e^{6t}|+C

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integral of (6e^{6t})/(9+e^{6t)}

\int \frac{6 e ^{6 t}}{9 + e ^{6 t}} d t

ln 9 + e^{6 t} + C

Solution steps

\int \frac{6 e ^{6 t}}{9 + 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}}{9 + 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 = 9 + e^{6 t}

= 6 \cdot \frac{1}{6} ln 9 + e^{6 t}

Simplify

6 \cdot \frac{1}{6} ln 9 + e^{6 t} : ln 9 + e^{6 t}

= ln 9 + e^{6 t}

Add a constant to the solution

= ln 9 + e^{6 t} + C

\frac{d y}{d x} = \frac{( y + x ^{2} + y ^{2} )}{x}

\frac{x - 2}{2 ( x - 1 ) ^{\frac{3}{2}}}

n = 1 \sum \infty (2 n + 1)^{- 5}

\frac{\partial}{\partial x} (ln (1 - x^{2}))

x \to 4 - lim (\frac{2}{x ^{2} - 16})

What is the integral of (6e^{6t})/(9+e^{6t)} ?
The integral of (6e^{6t})/(9+e^{6t)} is ln|9+e^{6t}|+C