If int {frac{{dx}}{{{x^3}{{left( {1 + {x^6}} | vedclass

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(B) Given integral $I = \int \frac{dx}{x^{3}(1+x^{6})^{2 / 3}}$.We can rewrite the integral by factoring out $x^6$ from the parenthesis:$I = \int \fra

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$ - \frac{1}{{2{x^3}}}$

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(B) Given integral $I = \int \frac{dx}{x^{3}(1+x^{6})^{2 / 3}}$.We can rewrite the integral by factoring out $x^6$ from the parenthesis:$I = \int \frac{dx}{x^{3} \cdot (x^6)^{2/3} (1 + x^{-6})^{2/3}} = \int \frac{dx}{x^{3} \cdot x^4 (1 + x^{-6})^{2/3}} = \int \frac{dx}{x^7 (1 + x^{-6})^{2/3}}$.Let $t = 1 + x^{-6}$. Then $dt = -6x^{-7} dx$,which implies $\frac{dx}{x^7} = -\frac{dt}{6}$.Substituting these into the integral:$I = \int -\frac{1}{6} t^{-2/3} dt = -\frac{1}{6} \cdot \frac{t^{1/3}}{1/3} + C = -\frac{1}{2} t^{1/3} + C$.Substituting back $t = 1 + x^{-6} = \frac{x^6+1}{x^6}$:$I = -\frac{1}{2} \left( \frac{1+x^6}{x^6} \right)^{1/3} + C = -\frac{1}{2} \cdot \frac{(1+x^6)^{1/3}}{x^2} + C$.Comparing this with the given form $xf(x)(1+x^6)^{1/3} + C$:$xf(x)(1+x^6)^{1/3} = -\frac{1}{2x^2} (1+x^6)^{1/3}$.Dividing both sides by $x(1+x^6)^{1/3}$:$f(x) = \frac{-1}{2x^3}$.

If $\int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^6}} \right)}^{2/3}}}} = xf\left( x \right){{\left( {1 + {x^6}} \right)}^{\frac{1}{3}}} + C} $ where $C$ is a constant of integration,then the function $f(x)$ is equal to

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