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in Definite Integrals by (487 points)
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Find:

Integrate ((x- 1)1/4 )/(x6) dx

\(\int\frac{{x^4-1}^{1/4}dx}{x^6}\)

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1 Answer

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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

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