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{d ^{2}}{d x ^{2}} (2 sec (x))

f (x) = \frac{4 x}{x ^{2} + 1}

\frac{d}{d x} ((3 x - 5)^{2})

x \to 0 lim (\frac{5}{x ^{2} + x} - \frac{5}{x})

\frac{1}{2 - x}, 0

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