int (x^2+5) / (x^3-x^2+x+3) dx Use partial fractions to find the indefinite integral

Indefinite integral are written in the form of int f(x) dx = F(x) +C
 where: f(x) as the integrand
           F(x) as the anti-derivative function 
           C  as the arbitrary constant known as constant of integration
To determine the indefinite integral of int (x^2+5)/(x^3-x^2+x+3) dx , we apply partial fraction decomposition to expand the integrand: f(x)=(x^2+5)/(x^3-x^2+x+3)
The pattern on setting up partial fractions will depend on the factors  of the  denominator. The factored form of x^3-x^2+x+3 =(x+1)(x^2-2x+3) .
For the linear factor (x+1) , we will have partial fraction: A/(x+1) .
For the quadratic factor (x^2-2x+3) , we will have partial fraction: (Bx+C)/(x^2-2x+3) .
The integrand becomes:
(x^2+5)/(x^3-x^2+x+3) =A/(x+1)+(Bx+C)/(x^2-2x+3)
Multiply both side by the LCD =(x+1)(x^2-2x+3) .
((x^2+5)/(x^3-x^2+x+3) )*(x+1)(x^2-2x+3)=(A/(x+1)+(Bx+C)/(x^2-2x+3))*(x+1)(x^2-2x+3)
x^2+5=A(x^2-2x+3)+(Bx+C)(x+1)
We apply zero-factor property on (x+1)(x^2-2x+3) to solve for values we can assign on x.
x+1 =0 then x=-1
x^2-2x+3=0 then x=1+-sqrt(2)i
To solve for A , we plug-in x=-1 :
(-1)^2+5=A((-1)^2-2*(-1)+3)+(B*(-1)+C)(-1+1)
1+5=A(1+2+3)+(-B+C)*0
6 = 6A
6/6= (6A)/6
A=1
To solve for C , plug-in A=1  and x=0 so that B*x becomes 0 :
0^2+5=1(0^2-2*0+3)+(B*0+C)(0+1)
0+5=1(0-0+3)+ (0+C)(1)
5 = 3 +C
C= 5-3
C =2 .
To solve for B , plug-in A=1 , C=2 , and x=1 :
1^2+5=1(1^2-2*1+3)+(B*1+2)(1+1)
1+5 = 1 (1-2+3)+(B+2)(2)
6 = 2 +2B+4
2B = 6-2-4
2B=0
(2B)/2 = 0/2
B =0
Plug-in A = 1 , B =0 , and C=2 , we get the partial fraction decomposition:
(x^2+5)/(x^3-x^2+x+3) =1/(x+1)+(0x+2)/(x^2-2x+3)
                      =1/(x+1)+2/(x^2-2x+3)
The integral becomes:
int(x^2+5)/(x^3-x^2+x+3) dx = int [1/(x+1)+2/(x^2-2x+3)] dx
Apply the basic integration property: int (u+v) dx = int (u) dx + int (v) dx
int [1/(x+1)+2/(x^2-2x+3)] dx =int 1/(x+1)dx +int 2/(x^2-2x+3)dx
For the first integral, we apply integration formula for logarithm: int 1/u du = ln|u|+C .
Let u =x+1 then du = dx
int 1/(x+1) dx =int 1/u du
                 = ln|u|
                 = ln|x+1|
Apply indefinite integration formula for rational function:
int 1/(ax^2+bx+c) dx = 2/sqrt(4ac-b^2)arctan((2ax+b)/sqrt(4ac-b^2)) +C
By comparing "ax^2 +bx +c " with "x^2-2x+3 ", we determine the corresponding values: a=1 , b=-2 , and c=3 .
The second integral becomes:
int 2/(x^2-2x+3)dx= 2int 1/(x^2-2x+3)dx
=2*[2/sqrt(4*1*3-(-2)^2)arctan((2*1x+(-2))/sqrt(4*1*3-(-2)^2))]
=2*[2/sqrt(12-4)arctan((2x-2)/sqrt(12-4))]
=2*[2/sqrt(8)arctan((2x-2)/sqrt(8))]
=2*[2/(2sqrt(2))arctan((2(x-1))/(2sqrt(2)))]
=2/sqrt(2)arctan((x-1)/sqrt(2))
=(2arctan((x-1)/sqrt(2))) /sqrt(2)
Combining the results, we get the indefinite integral as: 
int (x^2+5)/(x^3-x^2+x+3) dx =ln|x+1|+(2arctan((x-1)/sqrt(2))) /sqrt(2)+C
                                =ln|x+1|+ sqrt(2)arctan((sqrt(2)(x-1))/2) +C
                                =ln|x+1|+ sqrt(2)arctan((xsqrt(2)-sqrt(2))/2) +C