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

The solution to ln(u)-1/2 ln|1+u^2|+1/(2(1+u^2))+C is ln((usqrt(|1+u^2|))/(|1+u^2|))+1/(2(1+u^2))+C

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simplify ln(u)-1/2 ln|1+u^2|+1/(2(1+u^2))+C

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

\frac{u}{∣ 1 + u ^{2} ∣ ^{\frac{1}{2}}} : \frac{u ∣ 1 + u ^{2} ∣}{∣ 1 + u ^{2} ∣}

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

(- \frac{1}{3}) ln (\frac{2}{3})

ln (\frac{x ^{8} x ^{2} + 2}{( x + 2 ) ^{3}})

lo g_{4} (3 2 \cdot 5)

(\frac{1}{3} ln ∣ e^{x} + 1 ∣ - \frac{1}{3} ln ∣ e^{x} - 2 ∣) \cdot e^{2 x}

lo g_{4} (32)

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