Indefinite Integrals

2)y=\sqrt{1+2\sqrt{x-x^2}}
\rightarrow \left(\frac{y^2-1}{2}\right)^2=x-x^2
\rightarrow x=\frac{1}{2}+\frac{\sqrt{2y^2-y^4}}{2}
\int\sqrt{1+2\sqrt{x-x^2}}\,dx=xy-\int \frac{1}{2}+\frac{\sqrt{2y^2-y^4}}{2}\,dy

=xy-\frac{y}{2}-\int \frac{y\sqrt{2-y^2}}{2}\,dy

To solve this integral substitute 2-y² = z.
On solving you'll find that,

\int\sqrt{1+2\sqrt{x-x^2}}\,dx=xy-\frac{y}{2}+\frac{(2-y^2)^{3/2}}{6}+C

\sqrt{a+\sqrt{b+\sqrt{x}}}=y
\rightarrow x = [(y^2-a)^2-b]^2=
\rightarrow x = y^8+4a^2y^2+(a^2-b)^2-4ay^6-4ay^2(a^2-b)+2(a^2-b)y^4

Using the identity,

\int y\,dx=xy-\int x\,dy
\int \sqrt{a+\sqrt{b+\sqrt{x}}}\,dx
=xy-\frac{y^9}{9}-4a^2\frac{y^3}{3}-(a^2-b)^2y-4a\frac{y^7}{7}-4a(a^2-b)\frac{y^3}{3}-2(a^2-b)\frac{y^5}{5}