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

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

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integral of (e^θ)/(a+be^θ)

\int \frac{e ^{θ}}{a + b e ^{θ}} d θ

\frac{1}{b} ln a + b e^{θ} + C

Solution steps

\int \frac{e ^{θ}}{a + b e ^{θ}} d θ

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^{θ}

= \frac{1}{b} ln a + b e^{θ}

Add a constant to the solution

= \frac{1}{b} ln a + b e^{θ} + C

\int 4 - 3 (1 + x^{2})^{- 1} d x

\int \frac{6 ( 1 + x )}{x ( x ^{2} + 7 x + 6 )} d x

\frac{\partial}{\partial x} (cos^{4} (x^{3} y^{5}))

x \to - 8 lim (x^{2} - 3)

f (x) = 4 x^{3} - 8 x

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