What is the integral of (9e^{9x})/(5+e^{9x)} ?

The integral of (9e^{9x})/(5+e^{9x)} is ln|5+e^{9x}|+C

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integral of (9e^{9x})/(5+e^{9x)}

\int \frac{9 e ^{9 x}}{5 + e ^{9 x}} d x

ln 5 + e^{9 x} + C

Solution steps

\int \frac{9 e ^{9 x}}{5 + e ^{9 x}} d x

Take the constant out:

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

= 9 \cdot \int \frac{e ^{9 x}}{5 + e ^{9 x}} d x

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 = 5 + e^{9 x}

= 9 \cdot \frac{1}{9} ln 5 + e^{9 x}

Simplify

9 \cdot \frac{1}{9} ln 5 + e^{9 x} : ln 5 + e^{9 x}

= ln 5 + e^{9 x}

Add a constant to the solution

= ln 5 + e^{9 x} + C

\frac{y ^{^{'}} - e ^{- t} + 6}{y} = - 6, y (0) = 1

\frac{\partial}{\partial x} (5 sin (2 x) cos (5 y))

\int_{0}^{3} 3 x + 16 d x

y = ln ∣ sin (x) ∣

\frac{d}{d x} ((x^{2 - x})^{5} (x - 3 x^{2}))

What is the integral of (9e^{9x})/(5+e^{9x)} ?
The integral of (9e^{9x})/(5+e^{9x)} is ln|5+e^{9x}|+C