Question

Integrate the rational function:

`(x^3 + x + 1)/(x^2 -1)`

Answer 1

Let `I = int (x^3 + x + 1)/(x^2 - 1)  dx`

Since `(x^3 + x + 1)/(x^2 - 1)` is an improper fraction, we convert it into a proper fraction by long division method.

`x^2 - 1) overline (x^3 + x + 1)(x`
              x³ - x
            -     +       
                     2x + 1
`(x^3 + x + 1)/(x^2 -1) = Q + R/D`

∴ `(x^3 + x + 1)/(x^2 - 1) = x + (2x + 1)/(x^2 - 1)`        ....(i)

Now, 

`(2x + 1)/(x^2 - 1) = (2x + 1)/ ((x + 1)(x - 1))`

`= A/(x + 1) + B/(x - 1)`

⇒ 2x + 1 = A (x - 1) + B (x + 1)              ....(ii)

Putting x = -1 in (ii), we get

-2 + 1 = A (-1-1)

⇒ `A = (-1)/-2 = 1/2`

Putting x = 1 in (ii), we get

2 + 1 = B (1 + 1)

⇒ `B = 3/2`

∴ `(2x + 1)/(x^2 - 1) = 1/(2 (x + 1)) + 3/ (2 (x - 1))`             ....(iii)

From (i) and (iii),

`(x^3 + x + 1)/(x^2 - 1) = x + 1/ (2(x + 1)) + 3/ (2 (x - 1))`

∴ `int(x^3 + x + 1)/(x^2 - 1)  dx`

`= int x  dx + 1/2 int dx/ (x + 1) + 3/2 int dx/ (x - 1)`

`= x^2/2 + 1/2 log |x + 1| + 3/2  log |x - 1| + C`