If int frac{log (1+x^4)}{x^3} d x=f(x) log le | vedclass

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(B) We are given the integral $I = \int \frac{\log (1+x^4)}{x^3} d x$. Using integration by parts,let $u = \log (1+x^4)$ and $dv = x^{-3} dx$. Then $d

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$\frac{g(x)}{2}$

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(B) We are given the integral $I = \int \frac{\log (1+x^4)}{x^3} d x$. Using integration by parts,let $u = \log (1+x^4)$ and $dv = x^{-3} dx$. Then $du = \frac{4x^3}{1+x^4} dx$ and $v = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$. $I = -\frac{1}{2x^2} \log (1+x^4) - \int \left(-\frac{1}{2x^2}ight) \frac{4x^3}{1+x^4} dx = -\frac{1}{2x^2} \log (1+x^4) + \int \frac{2x}{1+x^4} dx$. Let $t = x^2$,then $dt = 2x dx$. $I = -\frac{1}{2x^2} \log (1+x^4) + \int \frac{dt}{1+t^2} = -\frac{1}{2x^2} \log (1+x^4) + 	an^{-1}(x^2) + c$. Comparing this with $f(x) \log \left(\frac{1}{g(x)}ight) + 	an^{-1}(h(x)) + c$,we have $f(x) = \frac{1}{2x^2}$,$g(x) = 1+x^4$,and $h(x) = x^2$. Now,$f\left(\frac{1}{x}ight) = \frac{1}{2(1/x)^2} = \frac{x^2}{2}$. Then $h(x) \left[f(x) + f\left(\frac{1}{x}ight)ight] = x^2 \left(\frac{1}{2x^2} + \frac{x^2}{2}ight) = \frac{1}{2} + \frac{x^4}{2} = \frac{1+x^4}{2} = \frac{g(x)}{2}$.

If $\int \frac{\log (1+x^4)}{x^3} d x=f(x) \log \left(\frac{1}{g(x)}\right)+\tan ^{-1}(h(x))+c$,then $h(x)\left[f(x)+f\left(\frac{1}{x}\right)\right]=$

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