*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.
