If int frac{x^4+1}{x^6+1} dx = A tan^{-1} x + | vedclass

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(A) We have, $\int \frac{x^4+1}{x^6+1} dx = A 	an^{-1} x + B 	an^{-1} x^3 + c$. Let $I = \int \frac{x^4+1}{x^6+1} dx = \int \frac{(x^4-x^2+1) + x^2}

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$\left(1, \frac{1}{3}\right)$

Vedclass · Answer

(A) We have, $\int \frac{x^4+1}{x^6+1} dx = A 	an^{-1} x + B 	an^{-1} x^3 + c$. Let $I = \int \frac{x^4+1}{x^6+1} dx = \int \frac{(x^4-x^2+1) + x^2}{(x^2)^3 + 1^3} dx$. Using the identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$, we have $x^6+1 = (x^2+1)(x^4-x^2+1)$. Thus, $I = \int \frac{x^4-x^2+1}{(x^2+1)(x^4-x^2+1)} dx + \int \frac{x^2}{x^6+1} dx$. $I = \int \frac{1}{x^2+1} dx + \int \frac{x^2}{(x^3)^2+1} dx$. $I = 	an^{-1} x + \int \frac{x^2}{(x^3)^2+1} dx$. Let $x^3 = t$, then $3x^2 dx = dt$, so $x^2 dx = \frac{1}{3} dt$. $I = 	an^{-1} x + \frac{1}{3} \int \frac{1}{t^2+1} dt = 	an^{-1} x + \frac{1}{3} 	an^{-1} t + c$. $I = 	an^{-1} x + \frac{1}{3} 	an^{-1} x^3 + c$. Comparing with $A 	an^{-1} x + B 	an^{-1} x^3 + c$, we get $A=1$ and $B=\frac{1}{3}$.

If $\int \frac{x^4+1}{x^6+1} dx = A \tan^{-1} x + B \tan^{-1} x^3 + c$, then $(A, B) =$

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