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 + \(\frac 12(x+1)+\frac 32(x-1)\)

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

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