Question

If `int (dx)/(4x^2 - 1)` = A log `((2x - 1)/(2x + 1))` + c, then A = ______.

Options

• 1

• `1/2`

• `1/3`

• `1/4`

Answer 1

If `int (dx)/(4x^2 - 1)` = A log `((2x - 1)/(2x + 1))` + c, then A = `\underlinebb(1/2)`.

Explanation:

`int (dx)/(4x^2 - 1)` = A log `((2x - 1)/(2x + 1))` + c

Factor the denominator

4x² − 1 = (2x − 1) (2x + 1)

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

Use partial fractions

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

Multiply both sides by (2x − 1) (2x + 1)

1 = A(2x + 1) + B(2x − 1)

1 = 2Ax + A + 2Bx − B

= (2A + 2B) x + (A − B)

Equating coefficients:

2A + 2B = 0 ⇒ A + B = 0

A − B = 1

From A + B = 0 ⇒ B = −A

A − (−A) = 1

⇒ 2A = 1

A = `1/2`