# 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