With the disk method, the volume is given by the integral
[tex]\displaystyle\pi\int_{x=0}^{x=5}\left(\frac1{\sqrt{1+x}}\right)^2\,\mathrm dx=\pi\int_0^5\frac{\mathrm dx}{1+x}[/tex]
[tex]=\displaystyle\pi\int_{t=1}^{t=6}\frac{\mathrm dt}t[/tex]
where the substitution [tex]t=x+1[/tex] was made, giving [tex]\mathrm dt=\mathrm dx[/tex].
[tex]=\displaystyle\pi\ln|t|\bigg|_{t=1}^{t=6}[/tex]
[tex]=\pi(\ln6-\ln1)[/tex]
[tex]=\pi\ln6[/tex]
