712.
Setting x + 1 = 1z
=> dx = - 1z2
The Integral readily evaluates to :
I = ∫ - dz √(54-(z-32)2
Which is easy!
\hspace{-16}(1)\;\;\mathbf{\int \frac{1}{2x\sqrt{1-x}\sqrt{2-x+\sqrt{1-x}}}dx}$\\\\\\ $(2)\;\;\mathbf{\int\frac{1}{(x+1).\sqrt{1+x-x^2}}dx}$
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3 Answers
711. 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.
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