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

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

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

simplify

ln ∣ x ∣ + \frac{1}{x} - \frac{1}{2 x ^{2}} - ln ∣ x + 1 ∣

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

Solution steps

ln ∣ x ∣ + \frac{1}{x} - \frac{1}{2 x ^{2}} - ln ∣ x + 1 ∣

Apply log rule:

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

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

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

- lo g_{10} (1.2 \cdot 1 0^{3})

ln (\frac{x ^{2} - 25}{x ^{4}})

\frac{ln ( a )}{2} - ln (b) - 3 ln (c)

ln (\frac{- 4 i}{1 - i})

ln (\frac{b ^{\frac{1}{2}}}{c ( 0.8 - c )})

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