How do you integrate #int (3x+7)/(x^4-16)dx# using partial fractions?

Question

Evelyn Agard · Accepted Answer

#1/32[13 ln|x-2| -ln|x+2| - 6 ln(x^2 + 4) -14 arctan (x/2)] + C#

Step 1: Partial fraction decomposition We begin be factor the denominator

#(3x+7)/(x^4-16)= (3x+7)/((x^2 +4)(x^2-4)) = (3x+7)/((x-2)(x+2)(x^2+4))#

#=A/(x-2)+B/(x-2) +(Cx+D)/(x^2 +4) " " " " " " " "color(red)((1))#

#3x+7=A(x+2)(x^2+4) +B(x-2)(x^2+4) " " " " " " " " + (Cx+D)(x^2-4)" " " " " "color(red)((2) " #

FOIL the partial fraction expression we have #A(x^3 +2x^2 + 4x+8)# #B(x^3 -2x^2 +4x-8)# #Cx^3 +Dx^2-4Cx -4D#

#0x^3 : " " A+B + C = 0 " " " " " (Eq. 3)# #0x^2: " " 2A-2B+D= 0 " " " " " (Eq. 4)# #3x: " " "4A+4B-4C= 3" " " " " " " (Eq. 5)# #7: " " " 8 A-8B -4D = 7" " " " " " " (Eq. 6)#

Step 2: Solve the system of equation above

Multiply 4 to #Eq.3# and add to #Eq.5# #4*(A+B+C= 0) hArr 4A +4B + 4C = 0#

#+ (4A+4B-4C= 3)# #color(red)(8 A + 8B = 3) " " " (Eq. 7) #

Multiply #(Eq. 4)# by 4 and add to #(Eq.6)# #4(2A- 2B+D= 0) hArr 8A - 8B +4D = 0#

+#8A-8B+4D= 7# #color(red)(16 A- 16B= 7) (Eq. 8)#

Multiply #(Eq.7)# by 2, add to #(Eq.8)# #2(8a+8b = 3) hArr 16A -16B= 6#

#+16A-16B= 7# #32A= 13 hArr A= 13/32 " " " " (8) #

Substitute into either #(Eq.7)# or #(Eq.8)# to solve for B #13/32+8B= 3 hArr B= -1/32 " " " " " " (9)#

Substitute (8) and (9) into Eq. 3 and Eq.4 to solve for C and D #C= -3/8 " " " " " (10)# #D=-7/8 " " " " " (11) #

Step 3: Rewrite the partial faction using (1) and (8) (9) (10)(11) #(3x+7)/(x^4-16) = 13/(32(x-2)) -(1/(32(x+2))) - (3x+7)/(8(x^2 +4)#

Step 4: Rewrite the integral #int((3x+7)/(x^4 -16))dx = 13/32 int( 1/(x-2)) dx -1/32 int (1/(x+2)) dx - 3/8*int( x/(x^2 +4)) dx -7/8 int (1/(x^2 +4)) dx #

Step 5: Start integrating #= 13/32 ln |x-2| -1/32 ln|x+2| -3/16 ln(x^2 +4) -7/8 arctan (x/2) +C#

#1/32[13 ln|x-2| -ln|x+2| - 6 ln(x^2 + 4) -14 arctan (x/2) ]+ C#

Sorry, this problem is really really long....

Please feel free to edit this, if you see any mistake.

Christopher Allers · Answer

undefined To integrate the given expression using partial fractions, first factor the denominator:

x^4 - 16 = (x^2 + 4)(x^2 - 4) = (x^2 + 4)(x + 2)(x - 2)

Now, express the integrand as a sum of partial fractions:

(3x + 7)/(x^4 - 16) = A/(x + 2) + B/(x - 2) + (Cx + D)/(x^2 + 4)

To find A, B, C, and D, multiply both sides by the denominator:

3x + 7 = A(x - 2)(x^2 + 4) + B(x + 2)(x^2 + 4) + (Cx + D)(x + 2)(x - 2)

Expand and simplify:

3x + 7 = A(x^3 + 4x - 2x^2 - 8) + B(x^3 + 4x + 2x^2 + 8) + (Cx + D)(x^2 - 4)

3x + 7 = A(x^3 - 2x^2 + 4x - 8) + B(x^3 + 2x^2 + 4x + 8) + C(x^3 - 4x) + D(x^2 - 4x)

Now, equate coefficients:

For x^3 term: 0 = A + B + C
For x^2 term: 3 = -2A + 2B + D
For x term: 1 = 4A + 4B - 4C
For constant term: 7 = -8A + 8B

Solve these equations to find the values of A, B, C, and D. Once you have these values, you can integrate each partial fraction separately. The integral of (3x + 7)/(x^4 - 16)dx will then be the sum of the integrals of the partial fractions.