Let \(I = \int ln \left(\frac x2 + 1\right) dx\)
Put \(\frac x2 + 1 = t\)
⇒ \(\frac{dx}{2} = dt\)
⇒ dx = 2dt
∴ \(I = 2\int ln\,\,t \,\,\,dt\)
\(= 2 (t\, ln\,\,t - t) + C\)
\(= 2\left(\frac x2 + 1\right) ln \left(\frac x2 + 1 \right) - 2 \left(\frac x 2 + 1\right)+C\)
\(= (x + 2)ln\left(\frac x2 + 1\right)-(x + 2)+C\)
