Answer
$$\frac{1}{4}\ln \left| {\frac{{{x^4}}}{{{x^4} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\left( {{x^4} + 1} \right)}}} dx \cr
& {\text{Multiply by }}\frac{{{x^3}}}{{{x^3}}} \cr
& = \int {\frac{{{x^3}}}{{{x^4}\left( {{x^4} + 1} \right)}}} dx \cr
& {\text{Let }}{x^4} = u,\,\,\,\,4{x^3}dx = du \cr
& = \frac{1}{4}\int {\frac{{du}}{{u\left( {u + 1} \right)}}} \cr
& {\text{Decompose }}\frac{1}{{u\left( {u + 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{1}{{u\left( {u + 1} \right)}} = \frac{A}{u} + \frac{B}{{u + 1}} \cr
& {\text{Multiply by }}u\left( {u + 1} \right){\text{ and simplify}} \cr
& 1 = A\left( {u + 1} \right) + Bu \cr
& {\text{if }}u = 0,\,\,\,A = 1 \cr
& {\text{if }}u = - 1,\,\,\,B = - 1 \cr
& \cr
& {\text{Replace the coefficients}} \cr
& \frac{1}{{u\left( {u + 1} \right)}} = \frac{1}{u} - \frac{1}{{u + 1}} \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \frac{1}{4}\int {\frac{{du}}{{u\left( {u + 1} \right)}}} = \frac{1}{4}\int {\left( {\frac{1}{u} - \frac{1}{{u + 1}}} \right)} du \cr
& {\text{Integrate}} \cr
& = \frac{1}{4}\ln \left| u \right| - \frac{1}{4}\ln \left| {u + 1} \right| + C \cr
& = \frac{1}{4}\ln \left| {\frac{u}{{u + 1}}} \right| + C \cr
& \cr
& {\text{Replace }}u = {x^4} \cr
& = \frac{1}{4}\ln \left| {\frac{{{x^4}}}{{{x^4} + 1}}} \right| + C \cr} $$
