If int frac{3x+1}{(x-1)^3(x+1)} dx = A cdot l | vedclass

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(A) Let the integrand be $\frac{3x+1}{(x-1)^3(x+1)} = \frac{a}{x+1} + \frac{b}{x-1} + \frac{c}{(x-1)^2} + \frac{d}{(x-1)^3}$.By partial fraction decom

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$\frac{-5}{4}$

Vedclass · Answer

(A) Let the integrand be $\frac{3x+1}{(x-1)^3(x+1)} = \frac{a}{x+1} + \frac{b}{x-1} + \frac{c}{(x-1)^2} + \frac{d}{(x-1)^3}$.By partial fraction decomposition,$3x+1 = a(x-1)^3 + b(x+1)(x-1)^2 + c(x+1)(x-1) + d(x+1)$.Setting $x=1$,we get $3(1)+1 = d(1+1) \implies 4 = 2d \implies d=2$.Setting $x=-1$,we get $3(-1)+1 = a(-1-1)^3 \implies -2 = -8a \implies a = \frac{1}{4}$.Comparing coefficients of $x^3$: $0 = a + b \implies b = -a = -\frac{1}{4}$.Comparing constant terms: $1 = a(-1) + b(1) + c(-1) + d(1) \implies 1 = -\frac{1}{4} - \frac{1}{4} - c + 2 \implies 1 = \frac{3}{2} - c \implies c = \frac{1}{2}$.Comparing with the given form,$A = a = \frac{1}{4}$,$B = c = \frac{1}{2}$,and $C = \frac{d}{2} = 1$ is incorrect; the form is $\frac{B}{x-1} + \frac{C}{(x-1)^2}$.Matching terms: $\int \frac{1/4}{x+1} dx + \int \frac{-1/4}{x-1} dx + \int \frac{1/2}{(x-1)^2} dx + \int \frac{2}{(x-1)^3} dx$.$= \frac{1}{4} \log|x+1| - \frac{1}{4} \log|x-1| - \frac{1/2}{x-1} - \frac{1}{(x-1)^2} + D$.$= \frac{1}{4} \log \left| \frac{x+1}{x-1} \right| - \frac{1/2}{x-1} - \frac{1}{(x-1)^2} + D$.Thus,$A = \frac{1}{4}$,$B = -\frac{1}{2}$,$C = -1$.$A+B+C = \frac{1}{4} - \frac{1}{2} - 1 = \frac{1-2-4}{4} = -\frac{5}{4}$.

If $\int \frac{3x+1}{(x-1)^3(x+1)} dx = A \cdot \log \left|\frac{x+1}{x-1}\right| + \frac{B}{x-1} + \frac{C}{(x-1)^2} + D$,then $A+B+C=$

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