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Question
Evaluate the following:
`int (x^2"d"x)/(x^4 - x^2 - 12)`
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Solution
Let I = `int (x^2"d"x)/(x^4 - x^2 - 12)`
= `int x^2/(x^4 - 4x^2 + 3x^2 - 12) "d"x`
= `int x^2/(x^2(x^2 - 4) + 3(x^2 - 4)) "d"x`
= `int x^2/((x^2 - 4)(x^2 + 3)) "d"x`
Put x2 = t for the purpose of partial fraction.
We get `"t"/(("t" - 4)("t" + 3))`
Let `"t"/(("t" - 4)("t" + 3)) = "A"/("t" - 4) + "B"/("t" + 3)` .....[where A and B are arbitrary constants]
`"t"/(("t" - 4)("t" + 3)) = ("A"("t" + 3) + "B"("t" - 4))/(("t" - 4)("t" + 3))`
⇒ t = At + 3A + Bt – 4B
Comparing the like terms, we get
A + B = 1 and 3A – 4B = 0
⇒ 3A = 4B
∴ A = `4/3 "B"`
Now `4/3 "B" + "B"` = 1
`7/3 "B"` = 1
∴ B = `3/7` and A = `4/3 xx 3/7 = 4/7`
So, A = `4/7` and B = `3/7`
∴ `int x^2/((x^2 - 4)(x^2 + 3)) "d"x`
= `4/7 int 1/(x^2 - 4) "d"x + 3/7 int 1/(x^2 + 3) "d"x`
= `4/7 int 1/(x^2 - (2)^2) "d"x + 3/7 int 1/(x^2 + (sqrt(3)^2) "d"x`
= `4/7 xx 1/(2 xx 2) log|(x - 2)/(x + 2)| + 3/7 xx 1/sqrt(3) tan^-1 x/sqrt(3)`
= `1/7 log |(x - 2)/(x + 2)| + sqrt(3)/7 tan^-1 x/sqrt(3) + "C"`
Hence, I = `1/7 log |(x - 2)/(x + 2)| + sqrt(3)/7 tan^-1 x/sqrt(3) + "C"`.
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