Evaluate $ 2\pi \int_0^1 \ln(x^2+2)\sqrt{1+\frac{4x^2}{(x^2+2)^2}}\,dx $ I've been trying to solve this this problem for so long but haven't had much luck, can someone please help me out?
The original question was:
Find the area of the part of the surface $z = \ln(x^2+y^2+2)$ that lies above the disk $x^2 + y^2 \le 1$ correct to four decimal places by expressing the area in terms of a single integral.
I set up the integral as:
$$
2\pi \int_0^1 \ln(x^2+2)\sqrt{1+\frac{4x^2}{(x^2+2)^2}}\,dx.
$$
edit 1: Sorry for being unclear, I need a numerical value at the end, I've tried solving it by parts but it hasn't gotten me anywhere. I could just put it into a calculator and get an answer but I'm more curious about how to actually set it up and the steps to solve this problem.

This is not a solid of revolution. The proble asks the area of the surface $z\log \left(x^2+y^2+2\right)$ inside the cylinder $x^2+y^2=1$
$z=\log \left(x^2+y^2+2\right)$ and $D:x^2+y^2\le 1$
$$S=\iint\limits_{D} \sqrt{z_x^2+z_y^2+1}\,dA$$
$$z_x=\frac{2 x}{x^2+y^2+2};\;z_y=\frac{2 y}{x^2+y^2+2}$$
$$S=\iint\limits_{D} \sqrt{z_x^2+z_y^2+1}\,dA$$
$$S=\iint\limits_{D}\sqrt{\frac{4 x^2}{\left(x^2+y^2+2\right)^2}+\frac{4 y^2}{\left(x^2+y^2+2\right)^2}+1}\,dA$$
$$S=\iint\limits_{D}\sqrt{1+\frac{4 \left(x^2+y^2\right)}{\left(x^2+y^2+2\right)^2}}\,dA$$
Changing to polar coordinates
$x=r\cos t;\;y=r\sin t$ and $Jacobian = r$ we get
$$S=\int _0^1\int _0^{2 \pi } r \sqrt{\frac{4 r^2}{\left(r^2+2\right)^2}+1}\,dtdr$$
$$S=\int _0^1\int _0^{2 \pi }\frac{r }{r^2+2}\sqrt{r^4+8 r^2+4}\,dtdr$$
$$S=\frac{1}{2} \pi  \left(2 \sqrt{13}-4+\log \left(\frac{1}{162} \left(95 \sqrt{13}+343\right)\right)-4 \sqrt{2} \cot ^{-1}\left(\sqrt{2}\right)+4 \sqrt{2} \cot ^{-1}\left(\sqrt{26}\right)\right)\approx 3.5618$$
