Question

How do you find `int 3/((1 + x)(1 - 2x))dx` using partial fractions?

Answer 1

`ln( (1 + x) / (1 - 2x)) + C`

Explanation:

Let `3 /( (1 + x) * (1 - 2x))` be = `(A /(1 + x) + B / (1 - 2x))`

Expanding the Right hand side, we get
`(A * (1 - 2x) + B * (1 + x)) / ((1 + x) * (1 - 2x)`
Equating, we get
`(A * (1 - 2x) + B * (1 + x)) / ((1 + x) * (1 - 2x)` = `3 /( (1 + x) * (1 - 2x))`

ie `A * (1 - 2x) + B * (1 + x) = 3`
or `A - 2Ax + B + Bx = 3`
or `(A + B) + x*(-2A + B) = 3`
equating the coefficient of x to 0 and equating constants, we get

`A + B = 3` and
`-2A + B = 0`
Solving for A & B, we get
`A = 1 and B = 2`
Substituting in the integration, we get
`int 3 /( (1 + x) * (1 - 2x))dx` = `int (1 /(1 + x) + 2 / (1 - 2x)) dx`

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

= `ln(1 + x) + 2 * ln(1 - 2x) * (-1 / 2)`

= `ln(1 + x) - ln(1 - 2x)`

= `ln((1 + x) /(1 - 2x)) + C`