[tex]\displaystyle\int_2^4\frac{1-\left(\frac x2\right)^2}x\,\mathrm dx[/tex]
Setting [tex]u=\dfrac x2[/tex] gives [tex]\mathrm du=\dfrac{\mathrm dx}2[/tex], or [tex]2\,\mathrm du=\mathrm dx[/tex]. This also makes [tex]2u=x[/tex].
When [tex]x\to2^+[/tex], you have [tex]u\to\dfrac22=1[/tex], while when [tex]x\to4^-[/tex], you have [tex]u\to\dfrac42=2[/tex].
Putting everything together, you find that
[tex]\displaystyle\int_2^4\frac{1-\left(\frac x2\right)^2}x\,\mathrm dx=\int_1^2\frac{1-u^2}{2u}(2\,\mathrm du)=\int_1^2\frac{1-u^2}u\,\mathrm du[/tex]
which means (B) is the correct answer.
