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

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Question

(B) Step $1$: Rewrite the numerator as $(x^4 - x^2 + 1) + x^2$. Step $2$: Split the integral into two parts: $I = \int \frac{x^4 - x^2 + 1}{x^6 + 1} d

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Step $1$: Rewrite the numerator as $(x^4 - x^2 + 1) + x^2$. Step $2$: Split the integral into two parts: $I = \int \frac{x^4 - x^2 + 1}{x^6 + 1} dx + \int \frac{x^2}{x^6 + 1} dx$. Step $3$: Note that $x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)$. Thus, the first integral becomes $\int \frac{1}{x^2 + 1} dx = 	an^{-1} x$. Step $4$: For the second integral, let $u = x^3$, then $du = 3x^2 dx$, so $x^2 dx = \frac{1}{3} du$. The integral becomes $\int \frac{1}{3(u^2 + 1)} du = \frac{1}{3} 	an^{-1}(u) = \frac{1}{3} 	an^{-1}(x^3)$. Step $5$: Combining both, 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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