Let's see if this can be rewritten.
#= int 1/sqrt(-2x^2 + x)dx#
#= int 1/sqrt(-2*(x^2 - x/2))dx#
Completing the Square:
#= int 1/sqrt(-2*(x^2 - x/2 + 1/16 - 1/16))dx#
#= int 1/sqrt(-2*(x-1/4)^2 + 1/8)dx#
u-substitution:
Now, let:
#u = x-1/4#
#du = dx#
#= int 1/sqrt(1/8-2u^2)du#
#= int 1/sqrt(1/8(1-16u^2))du#
#= int 1/(sqrt(1/8)sqrt(1-16u^2))du#
#= int 1/(sqrt(2)sqrt(1/16-u^2))du#
#= 1/sqrt2int 1/(sqrt(1/16-u^2))du#
Now that it looks better...
Trig substitution:
Let:
#u = asintheta# with #a = 1/4#,
with the form #sqrt(a^2 - x^2)# resembling #sqrt(1-sin^2theta)#.
#du = 1/4costhetad theta#
#sqrt(1/16 - u^2) = sqrt(1/4^2 - 1/4^2sin^2theta) = 1/4costheta#
We get:
#1/sqrt2int1/(cancel(1/4)cancel(costheta))1/cancel(4)cancel(costheta)d theta = theta/sqrt2 + C#
#= arcsin(4u)/sqrt2 + C#
but #u = x - 1/4#, so #4u = 4x - 1#:
#=> arcsin(4x - 1)/sqrt2 + C#
for #x in (0, 1/2)#
Since Wolfram Alpha does not give the same answer , I decided to check by taking the derivative of the integration result.
...or:
#d/(dx)[arcsin(4x - 1)/sqrt2 + C] = 1/sqrt2 * 1/(sqrt(1-(4x - 1)^2))*4#
#= 4/sqrt2*1/(sqrt(1-16x^2 + 8x - 1))#
#=(sqrt2)^3/(sqrt(-16x^2 + 8x)) = (sqrt8)/(sqrt(-16x^2 + 8x))#
#= (sqrt8)/(sqrt(-16x^2 + 8x))*(1/sqrt8)/(1/sqrt8)#
#= 1/(1/sqrt8sqrt(-16x^2 + 8x))#
#= 1/(sqrt(-(16/8)x^2 + (8/8)x))#
#= 1/(sqrt(-2x^2 + x)) = 1/(sqrt(x-2x^2)#
again, for #x in (0, 1/2)#
