The integral int frac{(x^8-x^2) dx}{(x^{12}+3 | vedclass

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Question

(A) Let $I = \int \frac{(x^8-x^2) dx}{(x^{12}+3x^6+1) 	an^{-1}(x^3+\frac{1}{x^3})}$.Divide the numerator and denominator by $x^6$:$I = \int \frac{(x^

Vedclass · Accepted Answer

$\log_e(|	an^{-1}(x^3+\frac{1}{x^3})|)^{1/3}+C$

Vedclass · Answer

(A) Let $I = \int \frac{(x^8-x^2) dx}{(x^{12}+3x^6+1) 	an^{-1}(x^3+\frac{1}{x^3})}$.Divide the numerator and denominator by $x^6$:$I = \int \frac{(x^2 - x^{-4}) dx}{(x^6 + 3 + x^{-6}) 	an^{-1}(x^3 + x^{-3})}$.Let $t = 	an^{-1}(x^3 + x^{-3})$.Then $dt = \frac{1}{1 + (x^3 + x^{-3})^2} \cdot (3x^2 - 3x^{-4}) dx$.$dt = \frac{3(x^2 - x^{-4})}{1 + x^6 + 2 + x^{-6}} dx = \frac{3(x^2 - x^{-4})}{x^6 + 3 + x^{-6}} dx$.Thus,$\frac{(x^2 - x^{-4}) dx}{x^6 + 3 + x^{-6}} = \frac{1}{3} dt$.Substituting this into the integral:$I = \int \frac{1}{3t} dt = \frac{1}{3} \ln |t| + C$.$I = \frac{1}{3} \ln |	an^{-1}(x^3 + x^{-3})| + C = \ln |	an^{-1}(x^3 + x^{-3})|^{1/3} + C$.Therefore,the correct option is $A$.

The integral $\int \frac{(x^8-x^2) dx}{(x^{12}+3x^6+1) \tan^{-1}(x^3+\frac{1}{x^3})}$ is equal to:

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