The value of int frac{1+x^{4}}{1+x^{6}} dx is | vedclass

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(B) We have,$I = \int \frac{1+x^{4}}{1+x^{6}} dx$. Divide the numerator and denominator by $x^2$: $I = \int \frac{\frac{1}{x^2} + x^2}{\frac{1}{x^2} +

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

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(B) We have,$I = \int \frac{1+x^{4}}{1+x^{6}} dx$. Divide the numerator and denominator by $x^2$: $I = \int \frac{\frac{1}{x^2} + x^2}{\frac{1}{x^2} + x^4} dx$. Alternatively,we can write: $\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}}{1+x^{6}} dx - \int \frac{x^{2}}{1+x^{6}} dx$. However,a simpler approach is: $\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx$ is not direct. Let's use the substitution method: $\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx$. Actually,the standard way is: $\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx$. Using the identity $1+x^6 = (1+x^2)(1-x^2+x^4)$: $\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{(1+x^{2})(1-x^{2}+x^{4})} dx = \int \frac{1}{1+x^2} dx + \int \frac{x^2(x^2-1)}{1+x^6} dx$. Following the provided steps: $\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx$ is complex. Correct approach: $\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx = 	an^{-1} x + \int \frac{x^2}{1+(x^3)^2} dx$. Let $t = x^3$,then $dt = 3x^2 dx$,so $x^2 dx = \frac{dt}{3}$. Thus,the integral becomes $	an^{-1} x + \frac{1}{3} \int \frac{dt}{1+t^2} = 	an^{-1} x + \frac{1}{3} 	an^{-1} t + C = 	an^{-1} x + \frac{1}{3} 	an^{-1} x^3 + C$.

The value of $\int \frac{1+x^{4}}{1+x^{6}} dx$ is

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