What is the integral of 1/2 ln((1+x)/(1-x)) ?

The integral of 1/2 ln((1+x)/(1-x)) is 1/2 (xln((1+x)/(1-x))+ln(|x+1||-x+1|))+C

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integral of 1/2 ln((1+x)/(1-x))

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

\frac{1}{2} (x ln (\frac{1 + x}{1 - x}) + ln (∣ x + 1 ∣ ∣ - x + 1 ∣)) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} (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 ∣)

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

Simplify

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

Add a constant to the solution

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

\int \frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{4}} d x

n = 1 \sum \infty \frac{1 0 ^{n}}{9 ^{n}}

x \frac{d y}{d x} - y = x^{2} e^{x}

\frac{\partial}{\partial x} (ln (3 x + 5 y))

x \to π lim ((2 x - 2 π) cot (x))

What is the integral of 1/2 ln((1+x)/(1-x)) ?
The integral of 1/2 ln((1+x)/(1-x)) is 1/2 (xln((1+x)/(1-x))+ln(|x+1||-x+1|))+C