The integral int {frac{{{x^7} + {x^2} + 1}}{{ | vedclass

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(D) Let $I = \int \frac{x^7 + x^2 + 1}{(3x^8 + 8x^3 + 24x)^{1/3}} dx$. Consider the substitution $t = 3x^8 + 8x^3 + 24x$. Then,differentiating with re

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$\frac{1}{{16}}{\left( {3{x^8} + 8{x^3} + 24x} \right)^{2/3}} + C$

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(D) Let $I = \int \frac{x^7 + x^2 + 1}{(3x^8 + 8x^3 + 24x)^{1/3}} dx$. Consider the substitution $t = 3x^8 + 8x^3 + 24x$. Then,differentiating with respect to $x$,we get $dt = (24x^7 + 24x^2 + 24) dx$. This simplifies to $dt = 24(x^7 + x^2 + 1) dx$,which implies $(x^7 + x^2 + 1) dx = \frac{1}{24} dt$. Substituting these into the integral,we have $I = \int \frac{1}{t^{1/3}} \cdot \frac{1}{24} dt$. $I = \frac{1}{24} \int t^{-1/3} dt = \frac{1}{24} \cdot \frac{t^{2/3}}{2/3} + C$. $I = \frac{1}{24} \cdot \frac{3}{2} t^{2/3} + C = \frac{1}{16} t^{2/3} + C$. Substituting back $t = 3x^8 + 8x^3 + 24x$,we get $I = \frac{1}{16} (3x^8 + 8x^3 + 24x)^{2/3} + C$.

The integral $\int {\frac{{{x^7} + {x^2} + 1}}{{{{\left( {3{x^8} + 8{x^3} + 24x} \right)}^{1/3}}}}dx} $ is equal to

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