By using the substitution $u = \sqrt{x+1}$, find $\displaystyle \int \frac{\sqrt{x+1}}{x-1}\,dx$. The region $R$ is bounded by the curve $y = \dfrac{\sqrt{x+1}}{x-1}$ and the lines $x = 8$ and $y = 1$. Find the exact area of $R$, simplifying your answer in the form $A - \sqrt{2}\,\ln\left(\dfrac{B-\sqrt{2}}{B+\sqrt{2}}\right)$ where $A$ and $B$ are integers to be determined. the volume of the solid generated when $R$ is rotated $2\pi$ radians about the x-axis, giving your answer correct to 2 decimal places.

1. Express $x$ in terms of $u$ and find $dx$ in terms of $du$.

$u = \sqrt{x+1} \implies x = u^2 - 1$
$dx = 2u \, du$

2. Substitute $x$ and $dx$ to express the integral entirely in terms of $u$.

$\int \frac{\sqrt{x+1}}{x-1}\,dx = \int \frac{u}{(u^2 - 1) - 1} (2u) \, du = \int \frac{2u^2}{u^2 - 2} \, du$

3. Simplify the improper rational fraction by algebraic manipulation or polynomial long division.

$\int \frac{2u^2}{u^2 - 2} \, du = \int \frac{2(u^2 - 2) + 4}{u^2 - 2} \, du = \int \left( 2 + \frac{4}{u^2 - 2} \right) \, du$

4. Integrate using the standard formula for $\frac{1}{x^2 - a^2}$.

$\int \left( 2 + \frac{4}{u^2 - 2} \right) \, du = 2u + 4 \left( \frac{1}{2\sqrt{2}} \ln \left| \frac{u-\sqrt{2}}{u+\sqrt{2}} \right| \right) + C$
$= 2u + \sqrt{2} \ln \left| \frac{u-\sqrt{2}}{u+\sqrt{2}} \right| + C$

5. Substitute $u = \sqrt{x+1}$ back to obtain the final answer in terms of $x$.

$2\sqrt{x+1} + \sqrt{2} \ln \left| \frac{\sqrt{x+1}-\sqrt{2}}{\sqrt{x+1}+\sqrt{2}} \right| + C$

1. Evaluate the indefinite integral from part (i) as it will be needed to calculate the area.

Let $u = \sqrt{x+1} \implies x = u^2 - 1 \implies dx = 2u\,du$.
$\int \frac{\sqrt{x+1}}{x-1}\,dx = \int \frac{u}{(u^2 - 2)} (2u)\,du = \int \frac{2u^2}{u^2 - 2}\,du$
$= \int \left( 2 + \frac{4}{u^2 - 2} \right) du = 2u + 4 \left( \frac{1}{2\sqrt{2}} \ln \left| \frac{u-\sqrt{2}}{u+\sqrt{2}} \right| \right)$
$= 2\sqrt{x+1} + \sqrt{2} \ln \left| \frac{\sqrt{x+1}-\sqrt{2}}{\sqrt{x+1}+\sqrt{2}} \right| + C$

2. Find the intersection of the curve $y = \frac{\sqrt{x+1}}{x-1}$ and the line $y = 1$ to determine the left boundary of the region.

$\frac{\sqrt{x+1}}{x-1} = 1 \implies \sqrt{x+1} = x - 1$
Since $x > 1$, squaring both sides gives:
$x + 1 = (x - 1)^2 = x^2 - 2x + 1 \implies x^2 - 3x = 0$
$x(x - 3) = 0 \implies x = 3$

3. Set up the definite integral for the area of the region $R$.

The region is bounded above by $y = 1$ and below by the curve between $x = 3$ and $x = 8$.
$\text{Area} = \int_{3}^{8} \left( 1 - \frac{\sqrt{x+1}}{x-1} \right) dx = [x]_3^8 - \int_{3}^{8} \frac{\sqrt{x+1}}{x-1}\,dx$
$= 5 - \int_{3}^{8} \frac{\sqrt{x+1}}{x-1}\,dx$

4. Evaluate the definite integral using the antiderivative.

$\int_{3}^{8} \frac{\sqrt{x+1}}{x-1}\,dx = \left[ 2\sqrt{x+1} + \sqrt{2} \ln \left| \frac{\sqrt{x+1}-\sqrt{2}}{\sqrt{x+1}+\sqrt{2}} \right| \right]_3^8$
Substitute $x = 8$: $2(3) + \sqrt{2} \ln \left( \frac{3-\sqrt{2}}{3+\sqrt{2}} \right) = 6 + \sqrt{2} \ln \left( \frac{3-\sqrt{2}}{3+\sqrt{2}} \right)$
Substitute $x = 3$: $2(2) + \sqrt{2} \ln \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right) = 4 + \sqrt{2} \ln \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right)$
Difference:
$2 + \sqrt{2} \left[ \ln \left( \frac{3-\sqrt{2}}{3+\sqrt{2}} \right) - \ln \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right) \right] = 2 + \sqrt{2} \ln \left( \frac{3-\sqrt{2}}{3+\sqrt{2}} \times \frac{2+\sqrt{2}}{2-\sqrt{2}} \right)$

5. Simplify the logarithmic argument and match the result to the required form to find $A$ and $B$.

Multiply the binomials inside the logarithm:
$\frac{(3-\sqrt{2})(2+\sqrt{2})}{(3+\sqrt{2})(2-\sqrt{2})} = \frac{6 + 3\sqrt{2} - 2\sqrt{2} - 2}{6 - 3\sqrt{2} + 2\sqrt{2} - 2} = \frac{4+\sqrt{2}}{4-\sqrt{2}}$
Substituting back into the total area expression:
$\text{Area} = 5 - \left( 2 + \sqrt{2} \ln \left( \frac{4+\sqrt{2}}{4-\sqrt{2}} \right) \right) = 3 - \sqrt{2} \ln \left( \frac{4+\sqrt{2}}{4-\sqrt{2}} \right)$
To match the target form $A - \sqrt{2} \ln \left( \frac{B-\sqrt{2}}{B+\sqrt{2}} \right)$, we multiply the numerator and denominator of the logarithmic argument by $-1$:
$3 - \sqrt{2} \ln \left( \frac{-4-\sqrt{2}}{-4+\sqrt{2}} \right)$
By comparison, $A = 3$ and $B = -4$.

1. Find the $x$-coordinate where the curve intersects the line $y = 1$ to determine the lower limit of integration.

$\frac{\sqrt{x+1}}{x-1} = 1$
$x + 1 = (x - 1)^2$
$x^2 - 3x = 0$
$x(x - 3) = 0$
Since $x > 1$ for the curve to be well-defined and positive, $x = 3$. The region $R$ is thus bounded horizontally between $x = 3$ and $x = 8$.

2. Formulate the definite integral for the volume of revolution of region $R$ about the $x$-axis.

The region $R$ is bounded above by the horizontal line $y = 1$ and below by the curve $y = \frac{\sqrt{x+1}}{x-1}$. The volume $V$ generated is the difference between the volumes of the two solids of revolution:
$V = \pi \int_{3}^{8} (1)^2 \,dx - \pi \int_{3}^{8} \left( \frac{\sqrt{x+1}}{x-1} \right)^2 \,dx$
$V = \pi \int_{3}^{8} \left[ 1 - \frac{x+1}{(x-1)^2} \right] \,dx$

3. Evaluate the volume integral numerically using a graphing calculator.

Since the question asks for an answer correct to 2 decimal places, we can evaluate the integral directly using a graphing calculator:
$V \approx \pi (3.03295)$
$V \approx 9.528$
$V = 9.53 \text{ (to 2 d.p.)}$