int frac{x^4+1}{1+x^6} dx = | vedclass

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(D) We need to evaluate the integral $I = \int \frac{x^4+1}{1+x^6} dx$.Split the numerator as $x^4+1 = (x^4 - x^2 + 1) + x^2$.Then,$I = \int \frac{x^4

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$	an^{-1} x + \frac{1}{3} 	an^{-1} x^3 + c$

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(D) We need to evaluate the integral $I = \int \frac{x^4+1}{1+x^6} dx$.Split the numerator as $x^4+1 = (x^4 - x^2 + 1) + x^2$.Then,$I = \int \frac{x^4 - x^2 + 1}{1+x^6} dx + \int \frac{x^2}{1+x^6} dx$.Let $I_1 = \int \frac{x^4 - x^2 + 1}{1+(x^2)^3} dx$ and $I_2 = \int \frac{x^2}{1+(x^3)^2} dx$.For $I_1$,use the identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$:$I_1 = \int \frac{x^4 - x^2 + 1}{(1+x^2)(x^4 - x^2 + 1)} dx = \int \frac{1}{1+x^2} dx = 	an^{-1} x$.For $I_2$,let $x^3 = t$,then $3x^2 dx = dt$,so $x^2 dx = \frac{dt}{3}$:$I_2 = \int \frac{1}{1+t^2} \cdot \frac{dt}{3} = \frac{1}{3} 	an^{-1} t = \frac{1}{3} 	an^{-1} (x^3)$.Combining these,$I = I_1 + I_2 = 	an^{-1} x + \frac{1}{3} 	an^{-1} (x^3) + C$.

$\int \frac{x^4+1}{1+x^6} dx =$

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