What is the integral of (9e^{9t})/(3+e^{9t)} ?

The integral of (9e^{9t})/(3+e^{9t)} is ln|3+e^{9t}|+C

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integral of (9e^{9t})/(3+e^{9t)}

\int \frac{9 e ^{9 t}}{3 + e ^{9 t}} d t

ln 3 + e^{9 t} + C

Solution steps

\int \frac{9 e ^{9 t}}{3 + 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}}{3 + 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 = 3 + e^{9 t}

= 9 \cdot \frac{1}{9} ln 3 + e^{9 t}

Simplify

9 \cdot \frac{1}{9} ln 3 + e^{9 t} : ln 3 + e^{9 t}

= ln 3 + e^{9 t}

Add a constant to the solution

= ln 3 + e^{9 t} + C

\frac{d}{d x} (1 - 5 x)

n = 1 \sum \infty \frac{3}{2 ^{n}}

\int_{- 1}^{1} 5 x d x

\frac{d}{d t} (- 2 e^{- 2 t})

x^{2} - 2, at x = 3

What is the integral of (9e^{9t})/(3+e^{9t)} ?
The integral of (9e^{9t})/(3+e^{9t)} is ln|3+e^{9t}|+C