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