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Question
Evaluate: `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx
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Solution
Let I = `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx
We perform actual division and express the result as:
`"Dividend"/"Divisor" = "Quotient" + "Remainder"/"Divisor"`
x - 1
`2"x"^2 - "x" - 10)overline(2"x"^3 - 3"x"^2 - 9"x" + 1)`
`2"x"^3 - "x"^2 - 10"x"`
(-) (+) (+)
`- 2"x"^2 + "x" + 1`
`- 2"x"^2 + "x" + 10`
(+) (-) (-)
- 9
∴ I = `int("x - 1" + (-9)/(2"x"^2 - "x" - 10))` dx
`= int "x" * "dx" - int 1 * "dx" - 9 int 1/(2"x"^2 - "x" - 10) "dx"`
Here 2x2 - x - 10
`= 2("x"^2 + 1/2"x" + 1/16 - 5 - 1/16)`
`= 2 [("x" - 1/4)^2 - 81/16]`
∴ I = `int "x" * "dx" - int 1 * "dx" - 9/2 int 1/(("x" - 1/4)^2 - (9/4)^2)`dx
`= "x"^2/2 - "x" - 9/2 * 1/(2 (9/4)) log |("x" - 1/4 - 9/4)/("x" - 1/4 + 9/4)| + "c"_1`
`= "x"^2/2 - "x" - log |("x" -5/2)/("x + 2")| + "c"_1`
`= "x"^2/2 - "x" - log|("2x" - 5)/(2("x + 2"))| + "c"_1`
`= "x"^2/2 - "x" + log|(2("x + 2"))/("2x" - 5)| + "c"_1`
`= "x"^2/2 - "x" + log |("x + 2")/("2x - 5")| + log 2 + "c"_1`
∴ I = `"x"^2/2 - "x" + log|("x + 2")/("2x - 5")| + "c" "where" "c" = "c"_1 + log 2`
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