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Evaluate: ∫2x3-3x2-9x+1/2x2-x-10 dx

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Question

Evaluate: `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx

Sum
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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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Chapter 5: Integration - MISCELLANEOUS EXERCISE - 5 [Page 139]

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Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 5 Integration
MISCELLANEOUS EXERCISE - 5 | Q IV. 5) ii) | Page 139

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