Answer
$\frac{1}{3}\sqrt {{t^6} - 5} - \frac{{\sqrt 5 }}{3}{\cos ^{ - 1}}\frac{{\sqrt 5 }}{{\left| {{t^3}} \right|}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {{t^6} - 5} }}{t}} dt; \cr
& {\text{Let }}u = {t^3} \Rightarrow du = 3{t^2}dt,{\text{ }}dt = \frac{{du}}{{3{t^2}}},{\text{ then}} \cr
& \int {\frac{{\sqrt {{t^6} - 5} }}{t}} dt = \int {\frac{{\sqrt {{u^2} - 5} }}{t}\left( {\frac{1}{{3{t^2}}}} \right)} du \cr
& = \int {\frac{{\sqrt {{u^2} - 5} }}{{3{t^3}}}} du \cr
& = \frac{1}{3}\int {\frac{{\sqrt {{u^2} - 5} }}{u}} du \cr
& {\text{Using the formula}} \cr
& \int {\frac{{\sqrt {{u^2} - {a^2}} }}{u}du = } \sqrt {{u^2} - {a^2}} - a{\cos ^{ - 1}}\frac{a}{{\left| u \right|}} + C \cr
& {\text{Therefore}}{\text{,}} \cr
& = \frac{1}{3}\int {\frac{{\sqrt {{u^2} - 5} }}{u}} du = \frac{1}{3}\sqrt {{u^2} - {{\left( {\sqrt 5 } \right)}^2}} - \frac{{\sqrt 5 }}{3}{\cos ^{ - 1}}\frac{{\sqrt 5 }}{{\left| u \right|}} + C \cr
& {\text{Simplifying}} \cr
& = \frac{1}{3}\sqrt {{u^2} - 5} - \frac{{\sqrt 5 }}{3}{\cos ^{ - 1}}\frac{{\sqrt 5 }}{{\left| u \right|}} + C \cr
& {\text{Write in terms of }}t,{\text{ }}u = {t^3},{\text{ then}} \cr
& = \frac{1}{3}\sqrt {{t^6} - 5} - \frac{{\sqrt 5 }}{3}{\cos ^{ - 1}}\frac{{\sqrt 5 }}{{\left| {{t^3}} \right|}} + C \cr} $$
