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

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(B) Let $ I = \int \frac{x^2+1}{x(x^2-1)} dx $. Divide the numerator and denominator by $ x^2 $: $ I = \int \frac{1 + \frac{1}{x^2}}{x(1 - \frac{1}{x^

Vedclass · Accepted Answer

$ \log \left| \frac{x^2-1}{x} \right| + c $,where $ c $ is a constant of integration.

Vedclass · Answer

(B) Let $ I = \int \frac{x^2+1}{x(x^2-1)} dx $. Divide the numerator and denominator by $ x^2 $: $ I = \int \frac{1 + \frac{1}{x^2}}{x(1 - \frac{1}{x^2})} dx $. Let $ t = x - \frac{1}{x} $. Then $ dt = (1 + \frac{1}{x^2}) dx $. However,the given integral is $ \int \frac{x^2+1}{x^3-x} dx $. Alternatively,using partial fractions: $ \frac{x^2+1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} $. $ x^2+1 = A(x^2-1) + Bx(x+1) + Cx(x-1) $. For $ x=0, 1 = -A \Rightarrow A = -1 $. For $ x=1, 2 = 2B \Rightarrow B = 1 $. For $ x=-1, 2 = 2C \Rightarrow C = 1 $. $ I = \int (-\frac{1}{x} + \frac{1}{x-1} + \frac{1}{x+1}) dx $. $ I = -\log |x| + \log |x-1| + \log |x+1| + c $. $ I = \log \left| \frac{(x-1)(x+1)}{x} \right| + c = \log \left| \frac{x^2-1}{x} \right| + c $.

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

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