int (x^2+3)/(xsqrt(x^2-4)) dx Find the indefinite integral

Recall indefinite integral follows int f(x) dx = F(x)+C
 where:
f(x) as the integrand
F(x) as the antiderivative of f(x)
C as the constant of integration.
 
 The given problem: int (x^2+3)/(xsqrt(x^2-4)) dx has an integrand of f(x)=(x^2+3)/(xsqrt(x^2-4)) .
Apply u-substitution on f(x) dx by letting u =x^2 then du = 2x dx or dx= (du)/(2x) :
int (x^2+3)/(xsqrt(x^2-4))dx =int (u+3)/(xsqrt(u-4))*(du)/(2x)
                        =int ((u+3)du)/(2x^2sqrt(u-4))
                         =int ((u+3)du)/(2usqrt(u-4))
Apply the basic integration property: int c*f(x) dx = c int f(x) dx :
int ((u+3)du)/(2usqrt(u-4))=(1/2)int ((u+3)du)/(usqrt(u-4))
Apply the basic integration property for sum:
int (u+v) dx = int (u) dx+int (v) dx.
(1/2)int ((u+3)du)/(usqrt(u-4))=(1/2) [int (udu)/(usqrt(u-4))+int (3du)/(usqrt(u-4))]
For the integration of theint (udu)/(usqrt(u-4)) , we can cancel out the u:
int (udu)/(usqrt(u-4))=int (du)/sqrt(u-4)
Let  v= u-4 then dv =du .
Apply the Law of exponents: sqrt(x)= x^1/2 and 1/x^n= x^-n ,  we get:
int (du)/sqrt(u-4)=int (dv)/sqrt(v)
                           
                          
Apply the Power Rule for integration: int x^n dx= x^(n+1)/(n+1)+C
int v^(-1/2)dv=v^((-1)/2+1)/((-1)/2+1) +C
                  = v^(1/2)/(1/2)
                  =v^(1/2)*(2/1)
                  = 2v^(1/2)  or   2sqrt(v)
With v= u-4  then 2sqrt(v) = 2sqrt(u-4) .
The integral becomes:
int (du)/sqrt(u-4)=2sqrt(u-4).
 
For the integration of int (3du)/(usqrt(u-4)) , we basic integration property: int c*f(x) dx = c int f(x)
int (3du)/(usqrt(u-4))=3int (du)/(usqrt(u-4))
Let: v= sqrt(u-4)
Then square both sides to get v^2=u-4 then v^2+4 =u
Applying implicit differentiation on v^2=u-4 , we get: 2vdv = du .
Plug-in du =2v dv ,  u=v^2+4 and v=sqrt(u-4) , we get:
3 int (du)/(usqrt(u-4))=3int (2vdv)/((v^2+4)*v)
                    =3int (2dv)/((v^2+4))
                    =3*2int (dv)/(v^2+4)
                    =6int (dv)/(v^2+4)
The integral part resembles the basic integration for inverse tangent function:
int (dx)/(x^2+a^2) = (1/a)arctan(u/a)+C
Then,
6int (dv)/(v^2+4) =6*(1/2)arctan(v/2)+C
                 =3arctan(v/2)+C
Plug-in v =sqrt(u-4) , we get:
int (3du)/(usqrt(u-4)) =3arctan(sqrt(u-4)/2)+C
 
Combining the results, we get:
(1/2) [int (udu)/(usqrt(u-4))+int (3du)/(usqrt(u-4))] = (1/2)*[2sqrt(u-4)+3arctan(sqrt(u-4)/2)]+C
=sqrt(u-4)+3/2arctan(sqrt(u-4)/2)+C
Plug-in u = x^2 to get the final answer:
int (x^2+3)/(xsqrt(x^2-4)) dx= sqrt(x^2-4)+3/2arctan(sqrt(x^2-4)/2)+C