What is the integral of ln((1+x)/(1-x)) ?

The integral of ln((1+x)/(1-x)) is xln((1+x)/(1-x))+ln(|x+1||-x+1|)+C

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integral of ln((1+x)/(1-x))

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

x ln (\frac{1 + x}{1 - x}) + ln (∣ x + 1 ∣ ∣ - x + 1 ∣) + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2 x}{( x + 1 ) ( - x + 1 )} d x = - ln (∣ x + 1 ∣ ∣ - x + 1 ∣)

= x ln (\frac{1 + x}{1 - x}) - (- ln (∣ x + 1 ∣ ∣ - x + 1 ∣))

Simplify

= x ln (\frac{1 + x}{1 - x}) + ln (∣ x + 1 ∣ ∣ - x + 1 ∣)

Add a constant to the solution

= x ln (\frac{1 + x}{1 - x}) + ln (∣ x + 1 ∣ ∣ - x + 1 ∣) + C

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

sin (x + 1)

t^{2} sin (t) \cdot e^{- 2 t}

\frac{d}{d x} ((1 + x y) e^{x y})

(3 x + 28)^{\frac{1}{2}}

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