Let I = \(\int\frac{{x^4-1}^{1/4}dx}{x^6}\)
\(\int\frac{x^4(1-\frac1{x^4})}{x^6}dx\)
= \(\int\cfrac{{(x^4(1-\frac1{x^4})}^{1/4}}{x^6}dx\)
= \(\int\cfrac{{x(1-\frac1{x^4})}^{1/4}}{x^6}dx\)
= \(\int\cfrac{{(x^4(1-\frac1{x^4})}^{1/4}}{x^5}dx\)
Let 1 - \(\frac1{x^4}=t^4\)
⇒ \(\frac4{x^5}dx\) = 4t3dt
⇒ \(\frac1{x^5}dx\) = t3dt
\(\therefore\) I = \(\int\)(t4)1/4.t3dt
= \(\int\)t4dt
= \(\frac{t^5}5+c\)
= \(\frac15(1-\frac1{x64})^{5/4}+c\) (\(\because\) t = (1 - \(\frac1{x^4}\))1/4)
Hence, \(\int\frac{(x^4-1)^{1/4}}{x^6}dx\) = \(\frac15(1-\frac1{x^4})^{5/4}\) + c