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Question
`int ("x + 2")/(2"x"^2 + 6"x" + 5)"dx" = "p" int (4"x" + 6)/(2"x"^2 + 6"x" + 5) "dx" + 1/2 int "dx"/(2"x"^2 + 6"x" + 5)`, then p = ?
Options
`1/3`
`1/2`
`1/4`
2
Solution
`1/4`
Explanation:
Let x + 2 = p `"d"/"dx" (2"x"^2 + 6"x" + 5) + "q"`
= p(4x + 6) + q
∴ x + 2 = 4px + 6p + q
∴ 4p = 1 and 6p + q = 2
∴ p = `1/4`
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