Integration help

Question

How do you integrate (x^4+81)/(x(x^2+9)^2)

Answer 1

Partial Fraction decomposition

x^4 + 81 / [x(x^2+9)^2] = A/x + Bx+C/(x^2+9) + (Dx+E)/(x^2+9)^2

Multiplies by LCD = x(x^2+9)^2

(x^4+81) = a(x^2+9)^2 + (Bx+C)(x)(x^2+9) + x(Dx+E)

= a (x^4 + 18x^2 + 81) + (Bx+C) (x^3 + 9x) + (Dx^2+ex)

= a x^4 + 18a x^2 + 81a + (Bx^4 + Cx^3 + 9bx^2 + 9cx) + Dx^2 + ex

= (A+b)x^4 + Cx^3 + (18a + 9b +d) x^2 + (9c+e)x + 81a

a+b = 1

c = 0

18a+9b + d = 0

9c+e = 0 ---> e = 0

81a = 81 ---> a = 1 ---> b = 0

d = -18

1/x + (-18x)/(x^2+9)^2

check :

[(x^2+9)^2 - 18x^2 ]/ x(x^2+9)^2 =

[ x^4 + 18x^2 + 18 - 18x^2] / x(x^2+9)^2

[ x^4 + 18]/x(x^2+9)^2

Yes, the partial fraction decomposition is correct

Integrates each piece:

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integral (1/x) = ln x

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-18 integral x/(x^2+9)^2 U = x^2+9 --> dU = 2x dx ---> (1/2) dU = x dx

-9 integral dU / U^2

-9integral U^(-2) dU

-9 -U^-1

9/ U

9/(x^2+9)

check by diff:

9(x^2+9)^(-1) =

-9(x^2 + 9)^(-2) (2x) =

-18x/(x^2+9)^2

yes, the anti-derivative is correct

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ln x + 9/(x^2+9) + c