Answer
$$ - \frac{{{{\tan }^{ - 1}}{x^3}}}{{3{x^3}}} + \frac{1}{6}\ln \left| {\frac{x}{{{{\left( {{x^6} + 1} \right)}^{1/6}}}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\tan }^{ - 1}}{x^3}}}{{{x^4}}}} dx \cr
& = \int {{x^{ - 4}}{{\tan }^{ - 1}}{x^3}} dx \cr
& {\text{substitute }}u = {x^3},{\text{ }}du = 3{x^2}dx,{\text{ }}dx = \frac{{du}}{{3{x^2}}} \cr
& = \int {{x^{ - 4}}{{\tan }^{ - 1}}u} \left( {\frac{{du}}{{3{x^2}}}} \right) = \frac{1}{3}\int {{x^{ - 6}}{{\tan }^{ - 1}}u} du \cr
& = \frac{1}{3}\int {{{\left( {{x^3}} \right)}^{ - 2}}{{\tan }^{ - 1}}u} du \cr
& = \frac{1}{3}\int {{u^{ - 2}}{{\tan }^{ - 1}}u} du \cr
& {\text{A matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}106 \cr
& \int {{u^n}{{\tan }^{ - 1}}u} du = \frac{1}{{n + 1}}\left( {{u^{n + 1}}{{\tan }^{ - 1}}u - \int {\frac{{{u^{n + 1}}}}{{{u^2} + 1}}du} } \right) \cr
& n = - 2 \cr
& = \frac{1}{3}\left( {\frac{1}{{ - 2 + 1}}} \right)\left( {{u^{ - 2 + 1}}{{\tan }^{ - 1}}u - \int {\frac{{{u^{ - 2 + 1}}}}{{{u^2} + 1}}du} } \right) \cr
& = - \frac{1}{3}\left( {{u^{ - 1}}{{\tan }^{ - 1}}u - \int {\frac{{{u^{ - 1}}}}{{{u^2} + 1}}du} } \right) \cr
& = - \frac{1}{3}{u^{ - 1}}{\tan ^{ - 1}}u + \frac{1}{3}\int {\frac{1}{{u\left( {{u^2} + 1} \right)}}du} \cr
& {\text{Formula 85 }}\int {\frac{{du}}{{u\left( {{u^2} + {a^2}} \right)}} = \frac{1}{{2{a^2}}}\ln \left| {\frac{{{u^2}}}{{{a^2} + {u^2}}}} \right|} + C \cr
& = - \frac{1}{3}{u^{ - 1}}{\tan ^{ - 1}}u + \frac{1}{{6{a^2}}}\ln \left| {\frac{{{u^2}}}{{{a^2} + {u^2}}}} \right| + C \cr
& = - \frac{1}{3}{u^{ - 1}}{\tan ^{ - 1}}u + \frac{1}{6}\ln \left| {\frac{{{u^2}}}{{{u^2} + 1}}} \right| + C \cr
& {\text{with }}u = {x^3} \cr
& = - \frac{1}{3}{\left( {{x^3}} \right)^{ - 1}}{\tan ^{ - 1}}{x^3} + \frac{1}{6}\ln \left| {\frac{{{{\left( {{x^3}} \right)}^2}}}{{{{\left( {{x^3}} \right)}^2} + 1}}} \right| + C \cr
& = - \frac{{{{\tan }^{ - 1}}{x^3}}}{{3{x^3}}} + \frac{1}{6}\ln \left| {\frac{{{x^6}}}{{{x^6} + 1}}} \right| + C \cr
& = - \frac{{{{\tan }^{ - 1}}{x^3}}}{{3{x^3}}} + \ln \left| {\frac{{{x^{6/6}}}}{{{{\left( {{x^6} + 1} \right)}^{1/6}}}}} \right| + C \cr
& = - \frac{{{{\tan }^{ - 1}}{x^3}}}{{3{x^3}}} + \frac{1}{6}\ln \left| {\frac{x}{{{{\left( {{x^6} + 1} \right)}^{1/6}}}}} \right| + C \cr} $$
