int frac{x^2 + 1}{x(x^2 - 1)} , dx is equal t | vedclass

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Question

(A) 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 - \frac{1}{x}}

Vedclass · Accepted Answer

$\log \left| \frac{x^2 - 1}{x} \right| + c$

Vedclass · Answer

(A) Let $I = \int \frac{x^2 + 1}{x(x^2 - 1)} \, dx$.Divide the numerator and denominator by $x^2$:$I = \int \frac{1 + \frac{1}{x^2}}{x - \frac{1}{x}} \, dx$.Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) \, dx$.Substituting these into the integral,we get:$I = \int \frac{1}{t} \, dt = \log |t| + c$.Substituting back $t = x - \frac{1}{x} = \frac{x^2 - 1}{x}$,we get:$I = \log \left| \frac{x^2 - 1}{x} \right| + c$.

$\int \frac{x^2 + 1}{x(x^2 - 1)} \, dx$ is equal to

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