Prerequisites :
#(1): int{f(x)}^n*f'(x)dx={f(x)}^(n+1)/(n+1)+C_1, (n!=-1)#.
#(2): intsqrt(a^2-x^2)dx=1/2[xsqrt(a^2-x^2)+a^2arc sin(x/a)]+C_2#.
Assuming that the Problem is to find the integral :
#I=intxsqrt(1+x-x^2)dx#.
The technic, in such cases, is to determine #m,n in RR#, such that,
#x=m*d/dx(1+x-x^2)+n, i.e., #
# x=m(1-2x)+n=(-2m)x+(m+n)#.
Comparing the respective co-efficients, we get, #-2m=1, or, #
# m=-1/2#.
Also, #m+n=0, &, m=-1/2 rArr n=1/2#.
So, we replace #x# by #-1/2*d/dx(1+x-x^2)+1/2#, and get,
#I=int[-1/2*d/dx(1+x-x^2)+1/2]sqrt(1+x-x^2)dx#,
#=-1/2int[(1+x-x^2)^(1/2)d/dx(1+x-x^2)dx#
#+1/2intsqrt(1+x-x^2)dx#,
#=-1/2*(1+x-x^2)^(1/2+1)/(1/2+1)#
#+1/2intsqrt{(sqrt5/2)^2-(x-1/2)^2}dx#,
#=-1/3(1+x-x^2)^(3/2)#
#+1/2[1/2{(x-1/2)sqrt(1+x-x^2)+5/4arc sin ((x-1/2)/(sqrt5/2))}#,
#rArr I=-1/3(1+x-x^2)^(3/2)+1/8(2x-1)sqrt(1+x-x^2)#
#+5/16arc sin((2x-1)/sqrt5)+C#.
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