What is the integral of (9e^x)/(e^x-1) ?

The integral of (9e^x)/(e^x-1) is 9ln|e^x-1|+C

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integral of (9e^x)/(e^x-1)

\int \frac{9 e ^{x}}{e ^{x} - 1} d x

9 ln ∣ e^{x} - 1 ∣ + C

Solution steps

\int \frac{9 e ^{x}}{e ^{x} - 1} 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 ^{x}}{e ^{x} - 1} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 ln ∣ u ∣

Substitute back

u = e^{x} - 1

= 9 ln ∣ e^{x} - 1 ∣

Add a constant to the solution

= 9 ln ∣ e^{x} - 1 ∣ + C

\frac{\partial}{\partial x} (x^{4} + 5 x^{2} y)

y^{^{''}} - 7 y^{^{'}} + 10 y = 0

\int \frac{4}{x ( ln ( x ) ) ^{2}} d x

x^{4} - 3 x^{2} + 9

\int x^{2} e^{- \frac{x}{2}} d x

What is the integral of (9e^x)/(e^x-1) ?
The integral of (9e^x)/(e^x-1) is 9ln|e^x-1|+C