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Question
Evaluate: `int (5"x"^2 + 20"x" + 6)/("x"^3 + 2"x"^2 + "x")` dx
Solution
Let I = `int (5"x"^2 + 20"x" + 6)/("x"^3 + 2"x"^2 + "x")` dx
`= int (5"x"^2 + 20"x" + 6)/("x"("x"^2 + 2"x" + 1))` dx
`= int (5"x"^2 + 20"x" + 6)/("x"("x + 1")^2)` dx
Let `(5"x"^2 + 20"x" + 6)/("x"("x + 1")^2) = "A"/"x" + "B"/"x + 1" + "C"/("x + 1")^2`
∴ 5x2 + 20x + 6 = A(x + 1)2 + B(x + 1)x + Cx ...(i)
Putting x = 0 in (i), we get
5(0) + 20(0) + 6 = A(1)2 + B(1)(0) + C(0)
∴ A = 6
Putting x = - 1 in (i), we get
5 (1) + 20(- 1) + 6 = A (0)+ B (0) (- 1) + C (-1)
∴ - 9 = - C
∴ C = 9
Putting x = 1 in (i), we get
5 (1) + 20 (1) + 6 = A (2)2 + B (2) (1) + C (1)
∴ 31 = 4A + 2B + C
∴ 31 = 4(6) + 2B + 9
∴ B = - 1
∴ `(5"x"^2 + 20"x" + 6)/("x"("x + 1")^2) = 6/"x" + (-1)/"x + 1" + 9/("x + 1")^2`
∴ I = `int [6/"x" + (- 1)/"x + 1" + 9/("x + 1")^2]` dx
`= 6 int 1/"x" "dx" - int 1/"x + 1" "dx" + 9 int ("x + 1")^-2` dx
`= 6 log |"x"| - log |"x + 1"| + 9("x + 1")^-1/(-1)` + c
∴ I = `6 log |"x"| - log |"x + 1"| - 9/("x + 1")` + c
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