Integrate \(x^2 + x – 1\over x^2 – 1\) with respect to x.

Solution :

\(\int\) \(x^2 + x – 1\over x^2 – 1\) dx = \(\int\) (\(x^2 – 1\over x^2 – 1\) + \(x\over x^2 – 1\))dx

= \(\int\) 1 dx + \(\int\) \(x\over x^2 – 1\)) dx

Let \(x^2 – 1\) = t  \(\implies\) 2x dx = dt

= x + \(\int\) \(dt\over 2t\)

= x + \(1\over 2\) \(\int\) \(1\over t\) dt

since, integration of \(1\over x\) = log x

= x + \(1\over 2\) log t + C

Substituing the value of t here, we get

= x + \(log(x^2 – 1)\over 2\) + C

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