Integral_5

$\begin{aligned} I & = \int_0^1 \frac {\ln t}{\sqrt{1-t}} dt & \small \color{#3D99F6} \text{By identity }\int_a^b f(x)\ dx = \int_a^b f(a+b-x)\ dx \\ & = \int_0^1 \frac {\ln (1-t)}{\sqrt t} dt & \small \color{#3D99F6} \text{Let }x^2 = t \implies 2x\ dx = dt \\ & = \int_0^1 2\ln (1-x^2)\ dx \\ & = 2 \int_0^1 \big(\ln (1+x) + \ln (1-x) \big)\ dx \\ & = 2\bigg[(1+x)\ln(1+x) - x - (1-x)\ln (1-x) - x\bigg]_0^1 & \small \color{#3D99F6} \text{Note that }\lim_{x \to 1}(1-x)\ln(1-x) = 0 \\ & = 2(2\ln 2 - 2) = 4(\ln 2-1) \end{aligned}$

Therefore, $a+b+c = 4+2+1 = \boxed 7$ .

call Beta function

$\begin{aligned} & \Beta(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt\\ & \frac{d}{dx}\Beta(x,y)=\frac{d}{dx}\int_0^1 t^{x-1}(1-t)^{y-1}dt\\ & \Beta(x,y)\left(\psi(x)-\psi(x+y)\right)=\int_0^1 t^{x-1}(1-t)^{y-1}\ln(t)dt\\ & \Beta(1,\frac{1}{2})\left(\psi(1)-\psi(\frac{3}{2})\right)=\int_0^1 (1-t)^{-\frac{1}{2}}\ln(t)dt & & \color{#D61F06} \text{put } x=1,y=\frac{1}{2}\\ & \int_0^1 \frac{\ln(t)}{\sqrt{1-t}}dt=2(2\ln(2)-2)=4\left ( \ln(2) - 1 \right ) & & \color{#D61F06} \beta(1,\frac{1}{2})=2,\psi(1)=-\gamma,\psi(\frac{3}{2})=-\gamma-2\ln(2)+2\\ \end{aligned}$

where $\color{#D61F06} \gamma$ is the Euler–Mascheroni constant .