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

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

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

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

ln 6 + e^{9 t} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{9 u} d u

= 9 \cdot \int \frac{1}{9 u} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 \cdot \frac{1}{9} ln ∣ u ∣

Substitute back

u = 6 + e^{9 t}

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

Simplify

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

= ln 6 + e^{9 t}

Add a constant to the solution

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

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