Question

$\int_{0}^{\frac{7}{2}} \sqrt{x+\frac{1}{\sqrt{x+\frac{1}{x+...}}}} dx$.

Answer 1

$\frac{14}{3} + \ln 2$

Answer 2

To evaluate the integral $\int_{0}^{\frac{7}{2}} \sqrt{x+\frac{1}{\sqrt{x+\frac{1}{x+...}}}} dx$, we first simplify the expression inside the integral.

Let $y = \sqrt{x+\frac{1}{\sqrt{x+\frac{1}{x+...}}}}$.
This expression can be written recursively as $y = \sqrt{x+\frac{1}{y}}$.

Now, we solve for $y$ in terms of $x$:
Square both sides: $y^2 = x + \frac{1}{y}$
Multiply by $y$ (since $y = \sqrt{\dots}$, $y$ must be positive, so $y \ne 0$):
$y^3 = xy + 1$
Rearrange the terms to express $x$ in terms of $y$:
$xy = y^3 - 1$
$x = \frac{y^3 - 1}{y}$
$x = y^2 - \frac{1}{y}$

Now we perform a change of variables for the integral from $x$ to $y$.
First, find $dx$ in terms of $dy$:
$dx = \frac{d}{dy}\left(y^2 - \frac{1}{y}\right) dy$
$dx = \left(2y - (-\frac{1}{y^2})\right) dy$
$dx = \left(2y + \frac{1}{y^2}\right) dy$

Next, change the limits of integration from $x$ to $y$:

1. Lower limit: When $x=0$.
   Substitute $x=0$ into $x = y^2 - \frac{1}{y}$:
   $0 = y^2 - \frac{1}{y}$
   $0 = \frac{y^3 - 1}{y}$
   Since $y \ne 0$, we must have $y^3 - 1 = 0$.
   $y^3 = 1$. Since $y$ must be positive, $y=1$.

2. Upper limit: When $x=\frac{7}{2}$.
   Substitute $x=\frac{7}{2}$ into $x = y^2 - \frac{1}{y}$:
   $\frac{7}{2} = y^2 - \frac{1}{y}$
   $\frac{7}{2} = \frac{y^3 - 1}{y}$
   $7y = 2(y^3 - 1)$
   $2y^3 - 7y - 2 = 0$
   We look for positive integer roots. Let $P(y) = 2y^3 - 7y - 2$.
   Test $y=1$: $P(1) = 2(1)^3 - 7(1) - 2 = 2 - 7 - 2 = -7 \ne 0$.
   Test $y=2$: $P(2) = 2(2)^3 - 7(2) - 2 = 2(8) - 14 - 2 = 16 - 14 - 2 = 0$.
   So, $y=2$ is a root. Since $x(y) = y^2 - 1/y$ is an increasing function for $y>0$ ($x'(y) = 2y + 1/y^2 > 0$), $y=2$ is the correct upper limit.

Now substitute $y$ for the integrand and $dx$ into the integral, with the new limits:
$\int_{0}^{\frac{7}{2}} \sqrt{x+\frac{1}{\sqrt{x+\frac{1}{x+...}}}} dx = \int_{x=0}^{x=\frac{7}{2}} y \, dx$
$= \int_{y=1}^{y=2} y \left(2y + \frac{1}{y^2}\right) dy$
$= \int_{1}^{2} \left(2y^2 + \frac{y}{y^2}\right) dy$
$= \int_{1}^{2} \left(2y^2 + \frac{1}{y}\right) dy$

Finally, evaluate the definite integral:
$= \left[ 2 \frac{y^3}{3} + \ln|y| \right]_{1}^{2}$
$= \left( 2 \frac{2^3}{3} + \ln(2) \right) - \left( 2 \frac{1^3}{3} + \ln(1) \right)$
$= \left( \frac{16}{3} + \ln(2) \right) - \left( \frac{2}{3} + 0 \right)$
$= \frac{16}{3} - \frac{2}{3} + \ln(2)$
$= \frac{14}{3} + \ln(2)$