The solution is
int (x^2+x)/((3x^2-1)(x+7)) dx=int ((Ax+B)/(3x^2-1)+C/(x+7)) dx∫x2+x(3x2−1)(x+7)dx=∫(Ax+B3x2−1+Cx+7)dx
Start with the needed partial fraction
(x^2+x)/((3x^2-1)(x+7))=(Ax+B)/(3x^2-1)+C/(x+7)x2+x(3x2−1)(x+7)=Ax+B3x2−1+Cx+7
Least Common Denominator = (3x^2-1)(x+7)(3x2−1)(x+7)
(x^2+x)/((3x^2-1)(x+7))=((Ax+B)(x+7)+C(3x^2-1))/((3x^2-1)(x+7))x2+x(3x2−1)(x+7)=(Ax+B)(x+7)+C(3x2−1)(3x2−1)(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))x2+x(3x2−1)(x+7)=Ax2+Bx+7Ax+7B+3Cx2−C(3x2−1)(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))1⋅x2+1⋅x+0⋅x0(3x2−1)(x+7)=(A+3C)x2+(B+7A)x+(7B−C)x0(3x2−1)(x+7)
The equations can now be established
A+3C=1" "A+3C=1 first equation
B+7A=1" "B+7A=1 second equation
7B-C=0" "7B−C=0 third equation
from the third
C=7BC=7B
substitute this in the first
A+3(7B)=1A+3(7B)=1
A+21B=1A+21B=1
A=1-21B" "A=1−21B fourth equation
from the second B=1-7AB=1−7A
and substitute in the fourth equation
A=1-21B" "A=1−21B fourth equation
A=1-21(1-7A)A=1−21(1−7A)
A=1-21+147AA=1−21+147A
A=20/146=10/73A=20146=1073
then B=1-7A=1-7(10/73)=3/73B=1−7A=1−7(1073)=373
B=3/73B=373
C=7B=7(3/73)=21/73C=7B=7(373)=2173
Do the integration now
int (x^2+x)/((3x^2-1)(x+7)) dx=int ((Ax+B)/(3x^2-1)+C/(x+7)) dx∫x2+x(3x2−1)(x+7)dx=∫(Ax+B3x2−1+Cx+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∫x2+x(3x2−1)(x+7)dx=∫⎛⎜⎝(1073)x+(373)3x2−1+2173x+7⎞⎟⎠dx
int (x^2+x)/((3x^2-1)(x+7)) dx∫x2+x(3x2−1)(x+7)dx
=int (10/73x)/(3x^2-1)dx+int (3/73)/(3x^2-1)dx+int (21/73)/(x+7) dx=∫1073x3x2−1dx+∫3733x2−1dx+∫2173x+7dx
int (x^2+x)/((3x^2-1)(x+7)) dx∫x2+x(3x2−1)(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=106(73)∫6x3x2−1dx+373∫13x2−1dx+2173∫1x+7dx
int (x^2+x)/((3x^2-1)(x+7)) dx∫x2+x(3x2−1)(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=5219∫6x3x2−1dx+√373∫√3(√3x)2−12dx+2173∫1x+7dx
Use the integration formulas
int (du)/u=ln u∫duu=lnu and int (du)/(u^2-a^2)=1/(2a)ln ((u-a)/(u+a))∫duu2−a2=12aln(u−au+a)
int (x^2+x)/((3x^2-1)(x+7)) dx∫x2+x(3x2−1)(x+7)dx
=5/219*ln(3x^2-1)+sqrt(3)/146ln((xsqrt3-1)/(xsqrt3+1))+21/73*ln(x+7)+C_0=5219⋅ln(3x2−1)+√3146ln(x√3−1x√3+1)+2173⋅ln(x+7)+C0
God bless....I hope the explanation is useful.
