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

The integral of 1/(1-x^2)*ln((1+x)/(1-x)) is 1/4 ln^2((1+x)/(1-x))+C

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

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

\frac{1}{4} ln^{2} (\frac{1 + x}{1 - x}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{2} d u

= \int \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

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

Substitute back

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

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

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( \frac{1 + x}{1 - x} )}{2} : \frac{1}{4} ln^{2} (\frac{1 + x}{1 - x})

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

Add a constant to the solution

= \frac{1}{4} ln^{2} (\frac{1 + x}{1 - x}) + C

f (x) = 5 x^{2}, at x = 1

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

\frac{1}{4 z - 5}

y = (x^{2} - 35)^{6}, at x = 6

f (x) = \frac{x}{x ^{2} - 2}

What is the integral of 1/(1-x^2)*ln((1+x)/(1-x)) ?
The integral of 1/(1-x^2)*ln((1+x)/(1-x)) is 1/4 ln^2((1+x)/(1-x))+C