some integrals

1. I set √(1-x) = t

So, - dx2√(1-x) = dt

Setting in the Integral We have,

I = ∫ - dt (1-t²) √(t²+t+1)

Set t = 1/z and we have,

I = ∫ z dz (z²-1) √(z²+z+1)

Split Numerator as 1 ≡ z + 1 - 1

We have,

I = ∫ dx (z-1) √(z²+z+1) - I

=> 2I = ∫ dz (z-1) √(z²+z+1)

Which is a known form and is evaluated.

I has too many substitutions but still It can be Solved.

2.

Setting x + 1 = 1z

=> dx = - 1z²

The Integral readily evaluates to :

I = ∫ - dz √(54-(z-32)²

Which is easy!