What is (1-x)(1/(1-x^2))-1/2 ln((1+x)/(1-x)) ?

The solution to (1-x)(1/(1-x^2))-1/2 ln((1+x)/(1-x)) is 1/(x+1)-1/2 (ln(1+x)-ln(1-x))

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simplify (1-x)(1/(1-x^2))-1/2 ln((1+x)/(1-x))

simplify

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

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

Solution steps

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

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

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

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

Remove parentheses:

(a) = a

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

Remove parentheses:

(a) = a

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

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

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

- lo g_{10} (4.88 \cdot 1 0^{- 6})

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

2000 ln (\frac{14000}{14000 - 2100 ( 8 )}) - 9.8 (8)

- lo g_{10} (4.6 \cdot 1 0^{- 8})

lo g_{4} (\frac{c}{16})

What is (1-x)(1/(1-x^2))-1/2 ln((1+x)/(1-x)) ?
The solution to (1-x)(1/(1-x^2))-1/2 ln((1+x)/(1-x)) is 1/(x+1)-1/2 (ln(1+x)-ln(1-x))