If int {frac{{{x^4} + 1}}{{x{{left( {{x^2} + | vedclass

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(C) Let $I = \int \frac{x^4 + 1}{x(x^2 + 1)^2} dx$. Add and subtract $x^2$ in the numerator: $I = \int \frac{x^4 + x^2 - x^2 + 1}{x(x^2 + 1)^2} dx$ $I

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$A = 1, B = 1$

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(C) Let $I = \int \frac{x^4 + 1}{x(x^2 + 1)^2} dx$. Add and subtract $x^2$ in the numerator: $I = \int \frac{x^4 + x^2 - x^2 + 1}{x(x^2 + 1)^2} dx$ $I = \int \frac{x^2(x^2 + 1) - (x^2 - 1)}{x(x^2 + 1)^2} dx$ $I = \int \frac{x^2(x^2 + 1)}{x(x^2 + 1)^2} dx - \int \frac{x^2 - 1}{x(x^2 + 1)^2} dx$ $I = \int \frac{x}{x^2 + 1} dx - \int \frac{x^2 - 1}{x(x^2 + 1)^2} dx$. For the second integral,let $u = x^2 + 1$,then $du = 2x dx$. This does not simplify easily. Alternatively,rewrite the integrand: $\frac{x^4 + 1}{x(x^2 + 1)^2} = \frac{x^4 + 2x^2 + 1 - 2x^2}{x(x^2 + 1)^2} = \frac{(x^2 + 1)^2 - 2x^2}{x(x^2 + 1)^2} = \frac{1}{x} - \frac{2x}{(x^2 + 1)^2}$. Now integrate: $I = \int \frac{1}{x} dx - \int \frac{2x}{(x^2 + 1)^2} dx$. Let $t = x^2 + 1$,then $dt = 2x dx$. $I = \ln |x| - \int t^{-2} dt = \ln |x| - (\frac{t^{-1}}{-1}) + c = \ln |x| + \frac{1}{x^2 + 1} + c$. Comparing this with $A \ln |x| + \frac{B}{1 + x^2} + c$,we get $A = 1$ and $B = 1$.

If $\int {\frac{{{x^4} + 1}}{{x{{\left( {{x^2} + 1} \right)}^2}}}} dx = A \ln |x| + \frac{B}{{1 + {x^2}}} + c$,where $c$ is the constant of integration,then:

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