Question

Evaluate  : `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`

Answer 1

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"`