Answer
$$\frac{1}{4}\ln \left| {\frac{{\sqrt {1 - {x^4}} - 1}}{{\sqrt {1 - {x^4}} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {1 - {x^4}} }}} \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}{u^2} = 1 - {x^4},\,\,\,2u\,du = - 4{x^3}dx,\,\,\,\,\,\frac{{udu}}{{ - 2{x^3}}} = dx \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{dx}}{{x\sqrt {1 - {x^4}} }}} = \int {\frac{{udu}}{{x\left( { - 2{x^3}} \right)u}}} \cr
& = - \frac{1}{2}\int {\frac{{du}}{{{x^4}}}} \cr
& = - \frac{1}{2}\int {\frac{{du}}{{1 - {u^2}}}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{\left( {u + 1} \right)\left( {u - 1} \right)}}} \cr
& {\text{Decompose }}\frac{{du}}{{\left( {u + 1} \right)\left( {u - 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{1}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = \frac{A}{{u + 1}} + \frac{B}{{u - 1}} \cr
& {\text{Multiply by }}\left( {u + 1} \right)\left( {u - 1} \right){\text{ and simplify}} \cr
& 1 = A\left( {u - 1} \right) + B\left( {u + 1} \right) \cr
& {\text{if }}u = - 1,\,\,\,A = - 1/2 \cr
& {\text{if }}u = 1,\,\,\,B = 1/2 \cr
& \cr
& {\text{Replace the coefficients}} \cr
& \frac{1}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = \frac{{ - 1/2}}{{u + 1}} + \frac{{1/2}}{{u - 1}} \cr
& \frac{1}{2}\int {\frac{{du}}{{\left( {u + 1} \right)\left( {u - 1} \right)}}} = \frac{1}{4}\int {\left( {\frac{{ - 1}}{{u + 1}} + \frac{1}{{u - 1}}} \right)} \cr
& \cr
& {\text{Integrate}} \cr
& = \frac{1}{4}\left( {\ln \left| {u - 1} \right| - \ln \left| {u + 1} \right|} \right) + C \cr
& = \frac{1}{4}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C \cr
& \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{4}\ln \left| {\frac{{\sqrt {1 - {x^4}} - 1}}{{\sqrt {1 - {x^4}} + 1}}} \right| + C \cr} $$
