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
