Question

Integrate the rational function:

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

Answer 1

Since `(1-x^2)/(x (1 - 2x)) = (1 - x^2)/(x - 2x^2)` is an improper fraction, therefore we convert it into a peoper fraction by long division method, we get

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

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

`= 1/2 int dx 1/2 int (x-2)/(2x^2 - x) dx`

Now, `(x - 2)/(2x^2 - x) = (x - 2)/(x (2x - 1))`

`= A/x + B/(2x - 1)`

⇒ x - 2 = A (2x - 1) + Bx                     ......(i)

Putting x = 0 in (i), we get

-2 = A (-1)

⇒ A = 2

Putting `x = 1/2` in (i), we get

`1/2 -2= B (1/2)`

⇒ 1 - 4 = B

⇒ B = -3

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

We have,

`int (1 - x^2)/(x (1 - 2x)) dx`

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

`= 1/2x + log |x| -3/4 log |1 - 2x| + C`