Question Detail

To find the integral of (\ln\left(\frac{x}{2}\right)\,dx), we use integration by parts, (\int u\,dv = uv - \int v\,du).Let (u = \ln\left(\frac{x}{2}\right)) and (dv = dx).Then (du = \frac{1}{x/2} \cdot \frac{1}{2}\,dx = \frac{1}{x}\,dx) and (v = x).So, (\int \ln\left(\frac{x}{2}\right)\,dx = x \ln\left(\frac{x}{2}\right) - \int x \cdot \frac{1}{x}\,dx)(= x \ln\left(\frac{x}{2}\right) - \int 1\,dx)(= x \ln\left(\frac{x}{2}\right) - x + C)(= x\left(\ln\left(\frac{x}{2}\right)-1\right) + C).