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Question
Evaluate : `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`
Solution
let I = `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`
Put x2 = t. to find constants A and B.
`"t"/("t"^2 + 5"t" + 6) = "t"/(("t" + 2)("t" + 3))`
`= "A"/("t" + 2) + "B"/"t + 3"`
∴ t = A (t + 3) + B (t + 2)
Putting t = -3 in equation (II).
-3 = B ( -1 ) ⇒ B = 3
Putting t = -2 in equation (II).
-2 = A (1) ⇒ A = -2
Substituting the values of A and replacing t by x2 in equation (I). we get
`"x"^2/("x"^4 + "5x"^2 + 6) = - 2/("x"^3 + 2) + 3/("x"^2 + 3)`
`therefore "I" = int [(-2)/("x"^2 + 2) + 3/("x"^2 + 3)] "dx"`
`= (-2) int "dx"/ ("x"^2 + 2) + 3 int "dx"/("x"^2 + 3)`
`= (-2) xx 1/sqrt2 "tan"^-1 ("x"/sqrt 2) + 3 xx 1/sqrt 3 "tan"^-1 ("x"/sqrt 3) + "c"`
`= -sqrt 2 "tan"^-1 ("x"/sqrt2) + sqrt 3 "tan"^-1 ("x"/sqrt 3) + "c"`
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