Answer
$S = \frac{49\pi}{3}$
Work Step by Step
The surface area of the line generated by revolving the curve around the x axis is given by:
$S = 2\pi\int_{a}^{b}y\sqrt {1+(\frac{dy}{dx})^{2}}dx$ where $y = \sqrt {x+1}, a= 1, b = 5 $
$\frac{dy}{dx} = \frac{1}{2\sqrt {x+1}}$
Hence
$S = 2\pi\int_{1}^{5} \sqrt {x+1} \sqrt {1+( \frac{1}{2\sqrt {x+1}})^{2}}dx$
or $S = 2\pi\int_{1}^{5} \sqrt {x+1} \sqrt {\frac{4x+5}{4(x+1)}}dx$
or $S = \pi\int_{1}^{5} \sqrt {4x+5}dx$
by means of substitution $ u = 4x+5 $
$du = 4dx$
$x=1 => u = 9$
$x=5 => u = 25$
$S = \frac{\pi}{4}\int_{9}^{25} \sqrt {u}du$
$S = \frac{\pi}{4}\frac{x^{3/2}}{3/2}$
$S = \frac{\pi}{4}\frac{25^{3/2}-9^{3/2}}{3/2}$
$S = \frac{49\pi}{3}$
