Evaluate using Partial Fraction Decomposition 3x2 x 7 x2 3x

Evaluate using Partial Fraction Decomposition. 3x2 +x +7 (x2 + 3)(x 2) 15. dx

!5) We have given integration of ((3x²+x+7)/((x²+3)(x-2)))dx

using partial fraction decomposition on (3x²+x+7)/((x²+3)(x-2))=(Ax+B)/(x^2+3)+C/(x-2)

integration of ((3x²+x+7)/((x²+3)(x-2)))dx=integration of ((Ax+B)/(x^2+3)+C/(x-2) )dx --(1)

(3x²+x+7)/((x²+3)(x-2))=[Ax^2+Bx-Ax-2B+Cx^2+3C]/((x^2+3)(x-2))

(3x²+x+7)/((x²+3)(x-2))=[(A+C)x^2+(B-A)x-2B+3C]/((x^2+3)(x-2))

set numerators equal both sides

(3x²+x+7)=(A+C)x^2+(B-A)x-2B+3C

setting the x^2 coefficient , x coefficient and constant terms on both sides

A+C=3

(B-A)=1

-2B+3C=7

A+C+B-A=3+1=4

B+C=4

from 2*(B+C=4)+(-2B+3C=7)

(2B+2C=8)+(-2B+3C=7)

5C=15

C=3

plug C=3 into A+C=3

A=3-C=3-3=0

plug A=0 into B-A=1

B=1+A=1+0=1

so we have A=0,B=1 and C=3

substitute A=0,B=1 and C=3 into equation (1)

integration of ((3x²+x+7)/((x²+3)(x-2)))dx=integration of ((0*x+1)/(x^2+3)+3/(x-2) )dx

=integration of (1/(x^2+3)+3/(x-2) )dx

=integration of (1/(x^2+3))dx+integration of (3/(x-2))dx

=integration of (1/(3*(x^2/3+1)))dx+integration of (3/(x-2))dx

substitute u=x/sqrt(3) and v=x-2

du/dx=1/sqrt(3) and dv/dx=1 implies dx=dv

dx=sqrt(3)*du and dx=dv

=integration of (sqrt(3)/(3*(u^2+1)))du+integration of (3/(v))dv

=sqrt(3)/3 *integration of (1/(u^2+1))du+3*ln(|v|)+C

=1/sqrt(3)*[arctan(u)]+3*ln(|v|)+C since sqrt(3)/3=sqrt(3)/(sqrt(3)*sqrt(3))=1/sqrt(3)

substitute back u=x/sqrt(3) and v=x-2

=1/sqrt(3)*[arctan(x/sqrt(3))]+3*ln(|x-2|)+C

=[arctan(x/sqrt(3))]/sqrt(3)+3*ln(|x-2|)+C

integration of ((3x²+x+7)/((x²+3)(x-2)))dx=[arctan(x/sqrt(3))]/sqrt(3)+3*ln(|x-2|)+C