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

The integral of (e^x)/(a+be^x) is 1/b ln|a+be^x|+C

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integral of (e^x)/(a+be^x)

\int \frac{e ^{x}}{a + b e ^{x}} d x

\frac{1}{b} ln ∣ a + b e^{x} ∣ + C

Solution steps

\int \frac{e ^{x}}{a + b e ^{x}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{b} ln ∣ u ∣

Substitute back

u = a + b e^{x}

= \frac{1}{b} ln ∣ a + b e^{x} ∣

Add a constant to the solution

= \frac{1}{b} ln ∣ a + b e^{x} ∣ + C

(3 + x^{2}) y^{^{'}} - 2 x y = 0

\int_{0}^{1} x x^{2} + 1 d x

\int (2 x^{3} - x^{2} + 3 x - 7) d x

\int_{0}^{2} 5 d x

\int \frac{1}{7 x ^{7}} d x

What is the integral of (e^x)/(a+be^x) ?
The integral of (e^x)/(a+be^x) is 1/b ln|a+be^x|+C