What is 1/(4(1+x^4))+ln(x)-1/4 ln(1+x^4) ?

The solution to 1/(4(1+x^4))+ln(x)-1/4 ln(1+x^4) is 1/(4(1+x^4))+ln((x(1+x^4)^{3/4})/(1+x^4))

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simplify 1/(4(1+x^4))+ln(x)-1/4 ln(1+x^4)

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

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

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

- lo g_{10} (3.16 \cdot 1 0^{- 8})

ln (4) + ln (8)

- lo g_{10} (3.5 \times 1 0^{- 4})

ln (\frac{x ^{\frac{2}{3}} y ^{6}}{w ^{3} z ^{4}})

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