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

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

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integral of (e^x)/(e^x+9)

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

ln ∣ e^{x} + 9 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = e^{x} + 9

= ln ∣ e^{x} + 9 ∣

Add a constant to the solution

= ln ∣ e^{x} + 9 ∣ + C

\int \frac{2 x ^{2}}{1 + x ^{2}} d x

\int - \frac{( x + 1 )}{x ^{2} - 1} d x

\int 9 cos (5 x) d x

\frac{d}{d x} (8 e^{6 x})

y^{^{'}} - 5 y = 2 e^{(5 x)}

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