If int frac{x}{left(x^2+1right)(x-1)} d x=A l | vedclass

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(C) Let $I = \int \frac{x}{(x^2+1)(x-1)} dx$.Using partial fractions, we write $\frac{x}{(x^2+1)(x-1)} = \frac{P}{x-1} + \frac{Qx+R}{x^2+1}$.Equating

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$\frac{3}{4}$

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(C) Let $I = \int \frac{x}{(x^2+1)(x-1)} dx$.Using partial fractions, we write $\frac{x}{(x^2+1)(x-1)} = \frac{P}{x-1} + \frac{Qx+R}{x^2+1}$.Equating the numerators: $x = P(x^2+1) + (Qx+R)(x-1) = (P+Q)x^2 + (R-Q)x + (P-R)$.Comparing coefficients: $P+Q=0$, $R-Q=1$, $P-R=0$.Solving these, we get $P = \frac{1}{2}$, $Q = -\frac{1}{2}$, $R = \frac{1}{2}$.Thus, $I = \int \frac{1/2}{x-1} dx + \int \frac{-1/2x + 1/2}{x^2+1} dx$.$I = \frac{1}{2} \log |x-1| - \frac{1}{4} \int \frac{2x}{x^2+1} dx + \frac{1}{2} \int \frac{1}{x^2+1} dx$.$I = \frac{1}{2} \log |x-1| - \frac{1}{4} \log |x^2+1| + \frac{1}{2} 	an^{-1} x + d$.Comparing with the given form, $A = -\frac{1}{4}$, $B = \frac{1}{2}$, $C = \frac{1}{2}$.Therefore, $A+B+C = -\frac{1}{4} + \frac{1}{2} + \frac{1}{2} = \frac{3}{4}$.

If $\int \frac{x}{\left(x^2+1\right)(x-1)} d x=A \log \left|x^2+1\right|+B \tan ^{-1} x+C \log |x-1|+d$, then $A+B+C=$

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