**Question 3** If the substitution, \( u = \frac{x}{2} \) is made, the integral \[ \int \left(\frac{1 - \frac{x}{2}}{x}\right)^2 \, dx \] is transformed into: - \(\int \frac{(1-u)^2}{2u} \, du\) - \(\int \frac{(1-u)^2}{4u} \, du\) - \(2 \int \frac{(1-u)^2}{u} \, du\) - \(\int \frac{(1-u)^2}{u} \, du\) --- Click "Save and Submit" to save and submit. Click "Save All Answers" to save all answers.

Consider the integral 

$\int \frac{{\left(1-\frac{x}{2}\right)}^2}{x}dx$

$\int \frac{{\left(1-\frac{x}{2}\right)}^2}{x}dx$

substitute 

$\begin{array}{l}u=\frac{x}{2}\\ \text{differentiate with respect to }x\\ \frac{du}{dx}=\frac{1}{2}\\ du=\frac{dx}{2}\\ 2du=dx\end{array}$

Now the integral becomes

$\begin{array}{l}\int \frac{{\left(1-\frac{x}{2}\right)}^2}{x}dx=\int \frac{{\left(1-u\right)}^2}{2u}\left(2du\right)\left(\because u=\frac{x}{2}\Rightarrow x=2u\right)\\ =\int \frac{{\left(1-u\right)}^2}{2u}\left(2du\right)\\ =\int \frac{{\left(1-u\right)}^2}{u}du\\ \int \frac{{\left(1-\frac{x}{2}\right)}^2}{x}dx=\int \frac{{\left(1-u\right)}^2}{u}du\end{array}$