Find int frac{(x^{4}-x)^{frac{1}{4}}}{x^{5}} | vedclass

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(A) We have $\int \frac{(x^{4}-x)^{\frac{1}{4}}}{x^{5}} dx = \int \frac{[x^{4}(1-\frac{1}{x^{3}})]^{\frac{1}{4}}}{x^{5}} dx = \int \frac{x(1-\frac{1}{

Vedclass · Accepted Answer

$\frac{4}{15}(1-\frac{1}{x^{3}})^{\frac{5}{4}}+C$

Vedclass · Answer

(A) We have $\int \frac{(x^{4}-x)^{\frac{1}{4}}}{x^{5}} dx = \int \frac{[x^{4}(1-\frac{1}{x^{3}})]^{\frac{1}{4}}}{x^{5}} dx = \int \frac{x(1-\frac{1}{x^{3}})^{\frac{1}{4}}}{x^{5}} dx = \int \frac{(1-\frac{1}{x^{3}})^{\frac{1}{4}}}{x^{4}} dx$. Let $t = 1 - \frac{1}{x^{3}} = 1 - x^{-3}$. Then $dt = -(-3)x^{-4} dx = \frac{3}{x^{4}} dx$,which implies $\frac{dx}{x^{4}} = \frac{dt}{3}$. Substituting these into the integral,we get $\int \frac{t^{\frac{1}{4}}}{3} dt = \frac{1}{3} \int t^{\frac{1}{4}} dt$. Integrating,we get $\frac{1}{3} 	imes \frac{t^{\frac{5}{4}}}{\frac{5}{4}} + C = \frac{1}{3} 	imes \frac{4}{5} t^{\frac{5}{4}} + C = \frac{4}{15} t^{\frac{5}{4}} + C$. Substituting back $t = 1 - \frac{1}{x^{3}}$,we get $\frac{4}{15}(1-\frac{1}{x^{3}})^{\frac{5}{4}} + C$.

Find $\int \frac{(x^{4}-x)^{\frac{1}{4}}}{x^{5}} dx$

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