If int frac{x^4+1}{x(x^2+1)^2} dx = A log |x| | vedclass

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(A) We have the integral $I = \int \frac{x^4+1}{x(x^2+1)^2} dx$. Multiply the numerator and denominator by $x$: $I = \int \frac{x(x^4+1)}{x^2(x^2+1)^2

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(A) We have the integral $I = \int \frac{x^4+1}{x(x^2+1)^2} dx$. Multiply the numerator and denominator by $x$: $I = \int \frac{x(x^4+1)}{x^2(x^2+1)^2} dx$. Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{dt}{2}$. Substituting this into the integral: $I = \frac{1}{2} \int \frac{t^2+1}{t(t+1)^2} dt$. Using partial fractions: $\frac{t^2+1}{t(t+1)^2} = \frac{A_1}{t} + \frac{B_1}{t+1} + \frac{C_1}{(t+1)^2}$. $t^2+1 = A_1(t+1)^2 + B_1 t(t+1) + C_1 t$. For $t=0$,$1 = A_1(1)^2 \implies A_1 = 1$. For $t=-1$,$2 = C_1(-1) \implies C_1 = -2$. Comparing coefficients of $t^2$: $1 = A_1 + B_1 \implies 1 = 1 + B_1 \implies B_1 = 0$. So,$I = \frac{1}{2} \int \left( \frac{1}{t} - \frac{2}{(t+1)^2} \right) dt = \frac{1}{2} \left( \log |t| + \frac{2}{t+1} \right) + c$. $I = \frac{1}{2} \log |x^2| + \frac{1}{x^2+1} + c = \log |x| + \frac{1}{x^2+1} + c$. Comparing with $A \log |x| + \frac{B}{1+x^2} + c$,we get $A = 1$ and $B = 1$. Thus,$A-B = 1-1 = 0$.

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

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