Question

`\rm\int_(0)^(2021) \frac{ \sqrt{x + \sqrt{x + \sqrt{x + √(x) + \cdots } } } }{1 + \sqrt{x - \sqrt{x - \sqrt{x - √(x - \cdots) } } } } dx \\`

Answer 1

We have the identity

`(a + b)^2 = a^2 + 2ab + b^2 = a^2 + ab + b (a + b) \\\\ ~~~~ \implies a + b = √(a^2 + ab + b (a + b)) \\\\ ~~~~ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b √(\cdots)}}}`

assuming
`a+b\ge0`.

For the numerator, let
`b=1`, so
`a\ge-1`, which means

`a^2 + a = x \implies a = \frac{-1 + √(1 + 4x)}2`

and so

`\sqrt{x + \sqrt{x + \sqrt{x + √(\cdots)}}} = \frac{-1 + √(1 + 4x)}2 + 1 = \frac{1 + √(1 + 4x)}2`

For the denominator, let
`b=-1` so that
`a\ge1`. Now

`a^2 - a = x \implies a = \frac{1 + √(1 + 4x)}2`

so that

`\sqrt{x - \sqrt{x - \sqrt{x - √(\cdots)}}} = \frac{1 + √(1 + 4x)}2 - 1 = \frac{-1 + √(1 + 4x)}2`

So, in the integral we can simplify and evaluate it to

`\displaystyle \int_0^(2021) \frac{\sqrt{x + \sqrt{x + \sqrt{x + √(\cdots)}}}}{1 + \sqrt{x - \sqrt{x - \sqrt{x - √(\cdots)}}}} \, dx \\\\ ~~~~~~~~ = \int_0^(2021) \frac{\frac{1 + √(1 + 4x)}2}{1 - \frac{1 - √(1+4x)}2} \, dx \\\\ ~~~~~~~~ = \int_0^(2021) (1 + √(1 + 4x))/(1 + √(1 + 4x)) \, dx \\\\ ~~~~~~~~ = \int_0^(2021) dx = \boxed{2021}`