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

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(D) We have the integral $I = \int \frac{x^4+1}{x^6+1} \, dx$. First,rewrite the numerator as $x^4+1 = (x^4-x^2+1) + x^2$. Since $x^6+1 = (x^2)^3 + 1^

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) We have the integral $I = \int \frac{x^4+1}{x^6+1} \, dx$. First,rewrite the numerator as $x^4+1 = (x^4-x^2+1) + x^2$. Since $x^6+1 = (x^2)^3 + 1^3 = (x^2+1)(x^4-x^2+1)$,we can split the integral: $\frac{x^4+1}{x^6+1} = \frac{x^4-x^2+1}{x^6+1} + \frac{x^2}{x^6+1} = \frac{1}{x^2+1} + \frac{x^2}{(x^3)^2+1}$. Now,integrate term by term: $\int \frac{1}{x^2+1} \, dx + \int \frac{x^2}{(x^3)^2+1} \, dx$. The first integral is $	an^{-1} x$. For the second integral,let $u = x^3$,then $du = 3x^2 \, dx$,so $x^2 \, dx = \frac{1}{3} du$. Thus,$\int \frac{1}{3} \frac{du}{u^2+1} = \frac{1}{3} 	an^{-1} u = \frac{1}{3} 	an^{-1} x^3$. Combining these,we get $	an^{-1} x + \frac{1}{3} 	an^{-1} x^3 + c$.

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

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