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)
