Rewrite the integrand as follows.
#int xsqrt(4x-x^2) "d"x = int xsqrt(4-(x-2)^2)"d"x#.
Denote the required integral #int xsqrt(4x-x^2) "d"x# by #I#. Make the substitution #u=x-2#. Then #I# is transformed to,
#I = int (u+2) sqrt(4-u^2) "d"u#.
#I = int usqrt(4-u^2) "d"u + 2int sqrt(4-u^2) "d"u#.
A good general integration result to be aware of is, for #n != -1#,
#int p'(x) [p(x)]^(n) "d"x = ([p(x)]^(n+1))/(n+1)+C#.
This can be easily verified by differentiating the LHS.
This can help us solve the first integral #int usqrt(4-u^2) "d"u# as picking #4-u^2=p(x)# gives #p'(x)=-2u#. We conclude, #int u(4-u^2)^(1/2) "d"u = -1/3 (4-u^2)^(3/2) + C#. Then,
#I = -1/3 (4-u^2)^(3/2) + 2intsqrt(4-u^2) "d"u#.
Denote #J=int sqrt(4-u^2) "d"u#. Make the substitution #u=2sin(theta)#. Then #"d"u = 2cos(theta) "d"theta#. #J# is transformed to,
#J = 2int sqrt(4(1-sin^2(theta)))cos(theta) "d"theta#.
The trigonometric identity #cos^2(theta)+sin^2(theta)=1# gives #sqrt(1-sin^2(theta))=cos^2(theta)#.
#J = 4 int cos^2(theta) "d"theta#.
The double angle identity #cos(2theta) = 2cos^2(theta)-1# gives #cos^2(theta)=1/2(1+cos(2theta))#. Then,
#J = 2 int 1+cos(2theta) "d"theta#
#J = 2theta + sin(2theta) + C#
#J = 2theta + 2sin(theta)cos(theta) + C#
#J = 2theta + 2sin(theta)sqrt(1-sin^2(theta))+C#.
Then #u=2sin(theta)# gives #theta = "arcsin"(u/2)#.
Then,
#J = 2"arcsin"(u/2) + usqrt(1-u^2/4)+C#
#J = 2"arcsin"(u/2) + u/2sqrt(4-u^2)+C#
We conclude,
#I = 4arcsin(u/2) + usqrt(4-u^2) - 1/3(4-u^2)^(3/2) + C#.
#I = 4arcsin(u/2) + 1/3(u^2+3u-4)sqrt(4-u^2) + C#.
Then #u=x-2#. #u^2=x^2-4x+4#. Gives #u^2+3u-4=x^2-x-6#.
Then,
#I = 2arcsin((x-2)/2) + 1/3 (x^2-x-6) sqrt(4x-x^2) + C#.
