int left[ frac{x^4-x}{x^{20}} right]^{1/4} dx | vedclass

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Question

(A) Let $I = \int \left( \frac{x^4-x}{x^{20}} \right)^{1/4} dx$. $I = \int \left( \frac{x^4(1 - 1/x^3)}{x^{20}} \right)^{1/4} dx = \int \frac{1}{x^4}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \left( \frac{x^4-x}{x^{20}} \right)^{1/4} dx$. $I = \int \left( \frac{x^4(1 - 1/x^3)}{x^{20}} \right)^{1/4} dx = \int \frac{1}{x^4} (1 - x^{-3})^{1/4} dx$. Let $1 - x^{-3} = t$. Then, differentiating both sides with respect to $x$, we get $3x^{-4} dx = dt$, which implies $\frac{1}{x^4} dx = \frac{1}{3} dt$. Substituting these into the integral: $I = \int (t)^{1/4} \cdot \frac{1}{3} dt = \frac{1}{3} \cdot \frac{t^{5/4}}{5/4} + C = \frac{4}{15} t^{5/4} + C$. Substituting $t = 1 - \frac{1}{x^3} = \frac{x^3-1}{x^3}$ back: $I = \frac{4}{15} \left( \frac{x^3-1}{x^3} \right)^{5/4} + C = \frac{4}{15} \left( \frac{(x^3-1)^5}{x^{15}} \right)^{1/4} + C$.

$\int \left[ \frac{x^4-x}{x^{20}} \right]^{1/4} dx =$

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