int frac{(x^4+x)^{frac{1}{4}}}{x^5} dx = . | vedclass

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Question

(A) To solve the integral $I = \int \frac{(x^4+x)^{\frac{1}{4}}}{x^5} dx$,we factor out $x^4$ from the expression inside the parenthesis: $I = \int \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To solve the integral $I = \int \frac{(x^4+x)^{\frac{1}{4}}}{x^5} dx$,we factor out $x^4$ from the expression inside the parenthesis: $I = \int \frac{(x^4(1+\frac{1}{x^3}))^{\frac{1}{4}}}{x^5} dx$ $I = \int \frac{x(1+\frac{1}{x^3})^{\frac{1}{4}}}{x^5} dx = \int \frac{(1+\frac{1}{x^3})^{\frac{1}{4}}}{x^4} dx$ Let $u = 1 + \frac{1}{x^3}$. Then $du = -\frac{3}{x^4} dx$,which implies $\frac{dx}{x^4} = -\frac{1}{3} du$. Substituting these into the integral: $I = \int u^{\frac{1}{4}} (-\frac{1}{3}) du = -\frac{1}{3} \int u^{\frac{1}{4}} du$ $I = -\frac{1}{3} \cdot \frac{u^{\frac{5}{4}}}{\frac{5}{4}} + C = -\frac{1}{3} \cdot \frac{4}{5} u^{\frac{5}{4}} + C$ $I = -\frac{4}{15} (1+\frac{1}{x^3})^{\frac{5}{4}} + C$. Thus,the correct option is $A$.

$\int \frac{(x^4+x)^{\frac{1}{4}}}{x^5} dx = $ . . . . . . $+ C$.

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