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प्रश्न
Evaluate the following:
`int (x^2 "d"x)/((x^2 + "a"^2)(x^2 + "b"^2))`
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उत्तर
Let I = `int (x^2 "d"x)/((x^2 + "a"^2)(x^2 + "b"^2))`
Put x2 = t for the purpose of partial fraction.
We get `"t"/(("t" + "a"^2)("t" + "b"^2))`
Put `"t"/(("t" + "a"^2)("t" + "b"^2)) = "A"/("T" + "a"^2) + "B"/("t" + "b"^2)`
⇒ `"t"/(("t" + "a"^2)("t" + "b"^2)) = ("A"("t" + "b"^2) + "B"("t" + "a"^2))/(("t" + "a"^2)("t" + "b"^2))`
⇒ t = At + Ab2 + Bt + Ba2
Comparing the like terms, we get
A + B = 1 and Ab2 + Ba2 = 0
A = `(-"a"^2)/"b"^2 "B"`
∴ `(-"a"^2)/"b"^2 "B" + "B"` = 1
`"B"((-"a"^2)/"b"^2 + 1)` = 1
⇒ `"B"((-"a"^2 + "b"^2)/"b"^2)` = 1
⇒ B = `"b"^2/("b"^2 - "a"^2)` and A = `(-"a"^2)/"b"^2 xx "b"^2/("b"^2 - "a"^2) = "a"^2/("a"^2 - "b"^2)`
So A = `"a"^2/("a"^2 - "b"^2)` and B = `(-"b"^2)/("a"^2 - "b"^2)`
∴ `int x^2/((x^2 + "a"^2)(x^2 + "b"^2)) "d"x = "a"^2/("a"^2 - "b"^2) int 1/(x^2 + "a"^2) "d"x - "b"^2/("a"^2 - "b"^2) int 1/(x^2 + "b"^2) "d"x`
= `"a"^2/("a"^2 - "b"^2) xx 1/"a" tan^-1 x/"a" - "b"^2/("a"^2 - "b"^2) * 1/"b" tan^-1 x/"b"`
= `"a"/("a"^2 - "b"^2) tan^-1 x/"a" - "b"/("a"^2 - "b"^2) tan^-1 x-"b" + "C"`
Hence, I = `1/("a"^2 - "b"^2) ["a" tan^-1 x/"a" - "b" tan^-1 x/"b"] + "C"`.
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