What is factor ln|y|-1/2 ln|y^2+1|+C ?

The solution to factor ln|y|-1/2 ln|y^2+1|+C is ln((|y|sqrt(|y^2+1|))/(|y^2+1|))+C

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factor ln|y|-1/2 ln|y^2+1|+C

factor

ln ∣ y ∣ - \frac{1}{2} ln y^{2} + 1 + C

ln (\frac{∣ y ∣ ∣ y ^{2} + 1 ∣}{∣ y ^{2} + 1 ∣}) + C

Solution steps

ln ∣ y ∣ - \frac{1}{2} ln y^{2} + 1 + C

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

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

Apply log rule:

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

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

= ln (\frac{∣ y ∣}{∣ y ^{2} + 1 ∣ ^{\frac{1}{2}}}) + C

\frac{∣ y ∣}{∣ y ^{2} + 1 ∣ ^{\frac{1}{2}}} = \frac{∣ y ∣ ∣ y ^{2} + 1 ∣}{∣ y ^{2} + 1 ∣}

= ln (\frac{∣ y ∣ ∣ y ^{2} + 1 ∣}{∣ y ^{2} + 1 ∣}) + C

- 5 ln (\frac{a}{c}) - 1 ln (\frac{c}{a})

lo g_{10} (0.1886) - lo g_{10} (0.2)

2 ln (e^{2 x} sin (x))

ln (0.8 e^{- x} x)

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

What is factor ln|y|-1/2 ln|y^2+1|+C ?
The solution to factor ln|y|-1/2 ln|y^2+1|+C is ln((|y|sqrt(|y^2+1|))/(|y^2+1|))+C