Psi Psi

$\psi(1+z) = -\gamma + \displaystyle \int_{0}^{1} \dfrac{1-x^{z}}{1-x}dx$
Putting $z = \dfrac{1}{2}$
$\psi\left(\dfrac{3}{2}\right) = -\gamma + \displaystyle \int_{0}^{1} \dfrac{1-\sqrt{x}}{1-x}dx = -\gamma + \displaystyle \int_{0}^{1} \dfrac{1}{1+\sqrt{x}}dx$
Put $x = t^{2} \longrightarrow dx = 2tdt$
$\therefore \psi\left(\dfrac{3}{2}\right) = -\gamma + \displaystyle 2\int_{0}^{1} \dfrac{t}{1+t}dt$
$\psi\left(\dfrac{3}{2}\right) = -\gamma + \displaystyle 2\int_{0}^{1} 1dt - \displaystyle 2\int_{0}^{1}\dfrac{1}{1+t}dt$
$\psi\left(\dfrac{3}{2}\right) = -\gamma + \left[2t\right]_{0}^{1} - \left[2\ln\left(1+t\right)\right]_{0}^{1}$
$\psi\left(\dfrac{3}{2}\right) = -\gamma + 2 - 2\ln(2)$

$\psi\left(\dfrac{3}{2}\right) = -\gamma + 2 - \ln(4)$

$A + B = 2+4 = 6$