The solution is
#int (x^2+x)/((3x^2-1)(x+7)) dx=int ((Ax+B)/(3x^2-1)+C/(x+7)) dx#
Start with the needed partial fraction
#(x^2+x)/((3x^2-1)(x+7))=(Ax+B)/(3x^2-1)+C/(x+7)#
Least Common Denominator = #(3x^2-1)(x+7)#
#(x^2+x)/((3x^2-1)(x+7))=((Ax+B)(x+7)+C(3x^2-1))/((3x^2-1)(x+7))#
Expanding the right side of the equation
#(x^2+x)/((3x^2-1)(x+7))=(Ax^2+Bx+7Ax+7B+3Cx^2-C)/((3x^2-1)(x+7))#
Match the numerical coefficients of the left and right sides of the equation
#(1*x^2+1*x+0*x^0)/((3x^2-1)(x+7))=((A+3C)x^2+(B+7A)x+(7B-C)x^0)/((3x^2-1)(x+7))#
The equations can now be established
#A+3C=1" "#first equation
#B+7A=1" "#second equation
#7B-C=0" "#third equation
from the third
#C=7B#
substitute this in the first
#A+3(7B)=1#
#A+21B=1#
#A=1-21B" "#fourth equation
from the second #B=1-7A#
and substitute in the fourth equation
#A=1-21B" "#fourth equation
#A=1-21(1-7A)#
#A=1-21+147A#
#A=20/146=10/73#
then #B=1-7A=1-7(10/73)=3/73#
#B=3/73#
#C=7B=7(3/73)=21/73#
Do the integration now
#int (x^2+x)/((3x^2-1)(x+7)) dx=int ((Ax+B)/(3x^2-1)+C/(x+7)) dx#
#int (x^2+x)/((3x^2-1)(x+7)) dx=int (((10/73)x+(3/73))/(3x^2-1)+(21/73)/(x+7)) dx#
#int (x^2+x)/((3x^2-1)(x+7)) dx#
#=int (10/73x)/(3x^2-1)dx+int (3/73)/(3x^2-1)dx+int (21/73)/(x+7) dx#
#int (x^2+x)/((3x^2-1)(x+7)) dx#
#=10/(6(73))int (6x)/(3x^2-1)dx+3/73int 1/(3x^2-1)dx+21/73int 1/(x+7) dx#
#int (x^2+x)/((3x^2-1)(x+7)) dx#
#=5/(219)int (6x)/(3x^2-1)dx+(sqrt(3))/73int (sqrt(3))/((sqrt(3)x)^2-1^2)dx+21/73int 1/(x+7) dx#
Use the integration formulas
#int (du)/u=ln u# and #int (du)/(u^2-a^2)=1/(2a)ln ((u-a)/(u+a))#
#int (x^2+x)/((3x^2-1)(x+7)) dx#
#=5/219*ln(3x^2-1)+sqrt(3)/146ln((xsqrt3-1)/(xsqrt3+1))+21/73*ln(x+7)+C_0#
God bless....I hope the explanation is useful.
