Rewrite
#x^2/(2x^2+7x+6) = 1/2(1 - (7x+6)/(2x^2+7x+6))#
Factor #2x^2+7x+6# as #(2x+3)(x+2)#. Perform a partial fraction decomposition to give
#x^2/(2x^2+7x+6) = 1/2 (1 - 8/(x+2) + 9/(2x+3))#
Consider the #n#th derivative of #(x+a)^(-1)#.
#"d"/("d"x) (x+a)^(-1) = (-1) * (x+a)^(-2)#
#"d"^2/("d"x^2) (x+a)^(-1) = (-1) * (-2) *(x+a)^(-3)#.
Easy to guess that #"d"^n/("d"x^n) (x+a)^(-n)= (-1)^(n)n!(x+a)^(-(n+1))#.
Proceed by induction. Assume for #n=k# that #"d"^k/("d"x^k) (x+a)^(-k) = (-1)^(k)k!(x+a)^(-(k+1))#.
Then,
#"d"^(k+1)/("d"x^(k+1)) (x+a)^(-k) = "d"/("d"x) ("d"^k/("d"x^k) (x+a)^(-k))#,
#"d"^(k+1)/("d"x^(k+1)) (x+a)^(-k) = "d"/("d"x) (-1)^(k)k!(x+a)^(-(k+1))#,
#"d"^(k+1)/("d"x^(k+1)) (x+a)^(-k) = (-1)^(k)k! (-(k+1)) (x+a)^(-(k+2))#,
#"d"^(k+1)/("d"x^(k+1)) (x+a)^(-k) = (-1)^(k+1)(k+1)! (x+a)^(-((k+1)+1))#,
Then if the expression holds for #n=k# this implies it holds for #n=k+1#. It is true for #n=1# thus it is true for all positive integer #n# by mathematical induction.
#"d"^n/("d"x^n) x^2/(2x^2+7x+6) = 1/2 "d"^(n)/("d"x^n) (1 - 8/(x+2) + 9/(2x+3))#
#"d"^n/("d"x^n) x^2/(2x^2+7x+6) = 1/2 "d"^(n)/("d"x^n) (1 - 8/(x+2) + (9/2)/(x+3/2))#
#"d"^n/("d"x^n) x^2/(2x^2+7x+6) = "d"^(n)/("d"x^n) (2) - 4 * "d"^(n)/("d"x^n) (x+2)^(-1) + 9/4 * "d"^(n)/("d"x^n) (x+3/2)^(-1)#
Our expression will hold for positive integer #n#. Clearly #"d"^n/("d"x^n) 2 = 0# for #n>=1#.
Then, substituting the result proved earlier,
#"d"^n/("d"x^n) x^2/(2x^2+7x+6) = 9/2^2(-1)^(n)n!(x+3/2)^(-(n+1)) - 4(-1)^(n)n!(x+2)^(-(n+1))#,
#"d"^n/("d"x^n) x^2/(2x^2+7x+6) = (-1)^(n)(n!)(9/(2^2(x+3/2)^(n+1)) - 4/((x+2)^(n+1)))#,
#"d"^n/("d"x^n) x^2/(2x^2+7x+6) = (-1)^(n)(n!)((9*2^(n-1))/((2x+3)^(n+1)) - 4/((x+2)^(n+1)))#.
