Question

Integrate the rational function: $\frac {x^3+x+1}{x^2-1}$

Answer 1

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (x3 + x + 1) by x2 − 1, we obtain

$\frac {x^3+x+1}{x^2-1}$ = x + $\frac {2x+1}{x^2-1}$

Let $\frac {2x+1}{x^2-1}$ = $\frac {A}{(x+1)}$ + $\frac {B}{(x-1)}$

$2x+1 = A(x-1)+B(x+1)$ .....(1)

Substituting x = 1 and −1 in equation (1), we obtain

A = $\frac 12$ and B = $\frac 32$

∴ $\frac {x^3+x+1}{x^2-1}$ = X + $\frac 12(x+1)+\frac 32(x-1)$

⇒ ∫$$\frac {x^3+x+1}{x^2-1}\ dx = $∫x dx +\frac 12 ∫\frac {1}{(x+1)}dx+\frac 32 ∫\frac {1}{(x-1)}dx$

                          = $\frac {x^2}{2}+\frac 12\ log|x+1|+\frac 32\ log|x-1|+C$