What is 4ln|x|-4ln|x+1|+3/x+C ?

The solution to 4ln|x|-4ln|x+1|+3/x+C is ln((x^4)/((x+1)^4))+3/x+C

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simplify 4ln|x|-4ln|x+1|+3/x+C

simplify

4 ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ + \frac{3}{x} + C

ln (\frac{x ^{4}}{( x + 1 ) ^{4}}) + \frac{3}{x} + C

Solution steps

4 ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ + \frac{3}{x} + C

Apply log rule:

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

4 ln ∣ x ∣ = ln (∣ x ∣^{4})

= ln (∣ x ∣^{4}) - 4 ln ∣ x + 1 ∣ + \frac{3}{x} + C

Apply log rule:

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

4 ln ∣ x + 1 ∣ = ln (∣ x + 1 ∣^{4})

= ln (∣ x ∣^{4}) - ln (∣ x + 1 ∣^{4}) + \frac{3}{x} + C

Apply log rule:

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

ln (∣ x ∣^{4}) - ln (∣ x + 1 ∣^{4}) = ln (\frac{∣ x ∣ ^{4}}{∣ x + 1 ∣ ^{4}})

= ln (\frac{∣ x ∣ ^{4}}{∣ x + 1 ∣ ^{4}}) + \frac{3}{x} + C

Simplify

\frac{∣ x ∣ ^{4}}{∣ x + 1 ∣ ^{4}} : \frac{x ^{4}}{( x + 1 ) ^{4}}

= ln (\frac{x ^{4}}{( x + 1 ) ^{4}}) + \frac{3}{x} + C

ln (\frac{1}{e ^{20}} - 1) - ln (20) (\frac{1}{2})

\frac{2 \cdot 0.053}{5} \cdot ln (\frac{0.45}{0.2})

- lo g_{10} (2.43 \cdot 1 0^{- 4})

\frac{ln ( 10 ) - ln ( 3 )}{2 ln ( 3 )}

ln (- α (x - θ)^{2})

What is 4ln|x|-4ln|x+1|+3/x+C ?
The solution to 4ln|x|-4ln|x+1|+3/x+C is ln((x^4)/((x+1)^4))+3/x+C