How do you integrate #int9/((x^2+9)(x+3)(x-3)) dx# using partial fractions?

Your partial fraction decomposition can be computed as follows:

Find #A#, #B#, #C# and #D# so that

#9 / ((x^2+9)(x+3)(x-3)) = (Ax + B)/(x^2+9) + C / (x+3) + D / (x-3)#

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To solve this equation for #A#, #B#, #C# and #D#, the first thing to do would be multiplying both sides with #(x^2+9)(x+3)(x-3)#:

#9 = (Ax + B)(x+3)(x-3) + C(x^2+9)(x-3) + D(x^2+9)(x+3)#

... expand the terms...

#9 = (Ax + B) (x^2 -9) + C (x^3 - 3x^2 + 9x - 27) + D(x^3 + 3x^2 + 9x + 27)#

#9 = Ax^3 + Bx^2 -9Ax - 9B + Cx^3 - 3Cx^2 + 9Cx - 27C + Dx^3 + 3Dx^2 + 9Dx + 27D#

... sort the #color(red)(x^3)# terms, #color(blue)(x^2)# terms, #color(violet)(x)# terms and terms #color(green)("without " x)#...

#color(red)(0 * x^3) + color(blue)(0 * x^2) + color(violet)(0 *x) + color(green)(9) = color(red)(Ax^3) + color(blue)(Bx^2) color(violet)(- 9Ax) color(green)(- 9B) + color(red)(Cx^3) color(blue)(- 3Cx^2) + color(violet)(9Cx) color(green)(- 27C) + color(red)(Dx^3) + color(blue)(3Dx^2) + color(violet)(9Dx) + color(green)(27D)#

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This equation can only hold if all the #x^3# terms match, all the #x^2# terms match, all the #x# terms match and all the terms without #x# match.

Thus, it can be split up into #4# equations:

# { ( (I) color(white)(xxx) 0 = color(white)(xx) A color(white)(xxxxxxi) + C color(white)(xxi)+ D color(white)(xxxxxxxx) color(red)(x^3) " terms" ), ( (II) color(white)(xx) 0 = color(white)(xxxxxx)B color(white)(xx)- 3C color(white)(xx) + 3D color(white)(xxxxxxiii) color(blue)(x^2) " terms" ), ( (III) color(white)(xii) 0 = -9A color(white)(xxxxxii) + 9C color(white)(xx)+ 9D color(white)(xxxxxxiii) color(violet)(x) " terms" ), ( (IV) color(white)(xx) 9 = color(white)(xxxx)-9B color(white)(x) -27C color(white)(x)+ 27D color(white)(xxx) color(green)("without " x )" terms") :}#

First of all, let's divide the equations #(III)# and #(IV)# by #9#:

# { ( (I) color(white)(xxx) 0 = color(white)(xx) A color(white)(xxxxxxi) + C color(white)(xxi)+ D ), ( (II) color(white)(xx) 0 = color(white)(xxxxxx)B color(white)(xx)- 3C color(white)(xx) + 3D), ( (III) color(white)(xii) 0 = -A color(white)(xxxxxxi) + C color(white)(xxi)+ D), ( (IV) color(white)(xx) 1 = color(white)(xxxx)-B color(white)(xx) -3C color(white)(xx)+ 3D) :}#

#(I) - (III)# gives

#0 = 2A " "=>" " A = 0#

#(II) - (IV)# gives

#-1 = 2B " "=>" " B = - 1 / 2#

Inserting #A = 0# in #(I)# and #B = - 1 /2# in #(II)# gives:

#{ ( (I') color(white)(xxx) 0 = color(white)(xx) C + D), ( (II') color(white)(xx) 1/2 = -3C + 3D):}#

From #(I')# we know that #C = -D#. Inserting this into #(II')# gives

#1/2 = 3D + 3D " " => " " D = 1/12#

and finally, #C = - 1 / 12#.

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Now that we have found #A#, #B#, #C# and #D#, we can apply the partial fraction decomposition and solve the integral:

# int 9 / ((x^2+9)(x+3)(x-3)) "d" x #

# = int (-1/2 * 1 / (x^2 +9) - 1/12 * 1/(x+3) + 1/12 * 1 /(x-3)) "d"x #

# = -1/2 int 1 / (x^2 +9) "d"x - 1/12 int 1/(x+3) "d"x + 1/12 int 1 /(x-3) "d"x #

Now, you need to compute three easier integrals instead of one complicated one.

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Since

#(arctan x)' = 1 / (x^2+1)#,

we know that

#[1/3 arctan (x/3)] ' = 1/3 * 1 /((x/3)^2 + 1) = 1/(3 (x^2/9 + 3)) = 1 / (x^2 + 9)#.

Also, we know that

#(ln abs(x+1))' = 1 / (x+1)#

Thus, we can solve the integral as follows:

# int 9 / ((x^2+9)(x+3)(x-3)) "d" x #

# = -1/2 int 1 / (x^2 +9) "d"x - 1/12 int 1/(x+3) "d"x + 1/12 int 1 /(x-3) "d"x #

# = - 1 /2 * 1 / 3 * arctan(x/3) - 1/12 ln abs(x+3) + 1/12 ln abs(x-3) + c#

# = - 1 /6 arctan(x/3) - 1/12 ln abs(x+3) + 1/12 ln abs(x-3) + c#