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

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(D) Let $I = \int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} $. We can write $x^5 dx$ as $x^3 \cdot x^2 dx$. Let $1 + x^3 = t^2$. Then $3x^2 dx = 2t dt$

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$\frac{2}{9}\sqrt {1 + {x^3}} ({x^3} - 2) + C$

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(D) Let $I = \int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} $. We can write $x^5 dx$ as $x^3 \cdot x^2 dx$. Let $1 + x^3 = t^2$. 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{{(t^2 - 1)}}{t} \cdot \frac{2}{3}t dt} = \frac{2}{3} \int {(t^2 - 1) dt} $. Integrating with respect to $t$: $I = \frac{2}{3} \left( \frac{t^3}{3} - t \right) + C = \frac{2}{9} t(t^2 - 3) + C$. Since $t = \sqrt{1 + x^3}$ and $t^2 = 1 + x^3$,we have: $I = \frac{2}{9} \sqrt{1 + x^3} (1 + x^3 - 3) + C = \frac{2}{9} \sqrt{1 + x^3} (x^3 - 2) + C$.

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

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