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Find:Integrate ((x4 - 1)1/4 )/(x6) dx\(\int\frac{{x^4-1}^{1/4}dx}{x^6}\) |
Answer» 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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