int frac{x^5}{sqrt{1 + x^3}} dx = | vedclass

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(C) Let $I = \int \frac{x^5}{\sqrt{1 + x^3}} dx$. Put $1 + x^3 = t^2$,then $3x^2 dx = 2t dt$,which implies $x^2 dx = \frac{2}{3} t dt$. Also,$x^3 = t^

Vedclass · Accepted Answer

$\frac{2}{9}(1 + x^3)^{3/2} - \frac{2}{3}(1 + x^3)^{1/2} + c$

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(C) Let $I = \int \frac{x^5}{\sqrt{1 + x^3}} dx$. Put $1 + x^3 = t^2$,then $3x^2 dx = 2t dt$,which implies $x^2 dx = \frac{2}{3} t dt$. Also,$x^3 = t^2 - 1$. Substituting these into the integral: $I = \int \frac{x^3 \cdot x^2 dx}{\sqrt{1 + x^3}} = \int \frac{(t^2 - 1) \cdot \frac{2}{3} t dt}{t}$ $I = \frac{2}{3} \int (t^2 - 1) dt$ $I = \frac{2}{3} \left( \frac{t^3}{3} - t \right) + c$ $I = \frac{2}{9} t^3 - \frac{2}{3} t + c$ Since $t = (1 + x^3)^{1/2}$,we have: $I = \frac{2}{9} (1 + x^3)^{3/2} - \frac{2}{3} (1 + x^3)^{1/2} + c$.

$\int \frac{x^5}{\sqrt{1 + x^3}} dx = $

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