int {frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}};d | vedclass

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Question

(A) We have $I = \int {\frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}}\,dx}$.Taking $x^4$ common from the bracket,we get $I = \int {\frac{{{{[x^4(1 - 1/x^3)]}

Vedclass · Accepted Answer

$\frac{4}{{15}}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$

Vedclass · Answer

(A) We have $I = \int {\frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}}\,dx}$.Taking $x^4$ common from the bracket,we get $I = \int {\frac{{{{[x^4(1 - 1/x^3)]}^{1/4}}}}{{{x^5}}}\,dx} = \int {\frac{{x{{(1 - 1/x^3)}^{1/4}}}}{{{x^5}}}\,dx} = \int {\frac{{{{(1 - 1/x^3)}^{1/4}}}}{{{x^4}}}\,dx}$.Let $1 - \frac{1}{{{x^3}}} = t$. Then,differentiating both sides with respect to $x$,we get $\frac{3}{{{x^4}}}dx = dt$,which implies $\frac{1}{{{x^4}}}dx = \frac{1}{3}dt$.Substituting these into the integral,we get $I = \int {{t^{1/4}} \cdot \frac{1}{3}dt} = \frac{1}{3} \cdot \frac{{{t^{5/4}}}}{{5/4}} + c = \frac{1}{3} \cdot \frac{4}{5} {t^{5/4}} + c = \frac{4}{{15}}{t^{5/4}} + c$.Substituting back $t = 1 - \frac{1}{{{x^3}}}$,we get $I = \frac{4}{{15}}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$.

$\int {\frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}}\;dx} $ is equal to

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