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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{1}{1 - x ^{2}} 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

\int_{0}^{\frac{π}{3}} sin (x) d x

y = ln (x + x^{2} + 9)

\frac{d}{d x} (3 (sin (x) + x cos (x)))

n = 1 \sum \infty 2^{\frac{1}{n}}

\frac{\partial}{\partial x} (3 x^{2} + 2 x y)

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