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Question
Integrate the following w.r.t. x : `(5x^2 + 20x + 6)/(x^3 + 2x ^2 + x)`
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Solution
Let I = `int (5x^2 + 20x + 6)/(x^3 + 2x^2 + x)*dx`
= `int (5x^2 + 20x + 6)/(x(x^2 + 2x + 1))*dx`
= `int (5x^2 + 20x + 6)/(x(x + 1)^2)*dx`
Let `(5x^2 + 20x + 6)/(x(x + 1)^2) = "A"/x + "B"/(x + 1) + "C"/(x + 1)^2`
∴ 5x2 + 20x + 6 = A(x + 1)2 + Bx(x + 1) + Cx
Put x = 0, we get
0 + 0 + 6 = A(1) + B(0)(1) + C(0)
∴ A = 6
Put x + 1 = 0, i x = – 1, we get
5(1) + 20(– 1) + 6 = A(0) + B(– 1)(0) + C(– 1)
∴ – 9 = – C
∴ C = 9
Put x = 1, we get
5(1) + 20(1) + 6 = A(4) + B(1)(2) + C(1)
But A = 6 and C = 9
∴ 31 = 24 + 2B + 9
∴ B = – 1
∴ `(5x^2 + 20x + 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`
= `6log|x| - log|x + 1| + 9*((x + 1)^-1)/(-1) + c`
= `log|x^6| - log|x + 1| - (9)/((x + 1)) + c`
= `log|x^6/(x + 1)| - (9)/((x + 1)) + c`.
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