int frac{1}{(x - 1)(x^2 + 1)} dx = | vedclass

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(A) We use partial fractions: $\frac{1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$.$1 = A(x^2 + 1) + (Bx + C)(x - 1)$.Setting $x =

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$\frac{1}{2}\log |x - 1| - \frac{1}{4}\log (x^2 + 1) - \frac{1}{2}	an^{-1} x + c$

Vedclass · Answer

(A) We use partial fractions: $\frac{1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$.$1 = A(x^2 + 1) + (Bx + C)(x - 1)$.Setting $x = 1$,we get $1 = A(1 + 1) \Rightarrow A = \frac{1}{2}$.Comparing coefficients of $x^2$: $A + B = 0 \Rightarrow B = -\frac{1}{2}$.Comparing constant terms: $A - C = 1 \Rightarrow C = A - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$.Thus,$\frac{1}{(x - 1)(x^2 + 1)} = \frac{1}{2(x - 1)} - \frac{x + 1}{2(x^2 + 1)}$.Integrating both sides: $\int \frac{1}{2(x - 1)} dx - \frac{1}{2} \int \frac{x}{x^2 + 1} dx - \frac{1}{2} \int \frac{1}{x^2 + 1} dx$.$= \frac{1}{2}\log |x - 1| - \frac{1}{4}\log (x^2 + 1) - \frac{1}{2}	an^{-1} x + c$.

$\int \frac{1}{(x - 1)(x^2 + 1)} dx = $

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