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Question
If `int (dx)/(4x^2 - 1)` = A log `((2x - 1)/(2x + 1))` + c, then A = ______.
Options
1
`1/2`
`1/3`
`1/4`
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Solution
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`
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