Integrate x root 1+ x - x^2 ?

Question

Find the integral : #intxsqrt(1+x-x^2)dx#.

Ezra Adams · Accepted Answer

# -1/3(1+x-x^2)^(3/2)+1/8(2x-1)sqrt(1+x-x^2)#

#+5/16arc sin((2x-1)/sqrt5)+C#.

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#. Feel & Spread the Joy of Maths.!

Luke Alvarez · Answer

undefined To integrate $ x\sqrt{1 + x - x^2} $, you can use the method of substitution. Let $ u = 1 + x - x^2 $. Then, $ du/dx = 1 - 2x $, and solving for $ dx $ gives $ dx = du / (1 - 2x) $. After substitution, integrate with respect to $ u $.