Integral Integral6

$\begin{aligned} I & = \int_0^1 \log \left(\frac 1{x^3}+1\right) dx \\ & = \int_0^1 \log \left(\frac {x^3+1}{x^3} \right) dx \\ & = \int_0^1 \left(\log (x^3+1) - 3\log x \right) dx \\ & = {\color{#3D99F6} \int_0^1 \log (x^3+1)\ dx} - 3 \int_0^1 \log x\ dx & \small \color{#3D99F6} \text{By integration by parts.} \\ & = {\color{#3D99F6} x\log(x^3+1)\bigg|_0^1 - \int_0^1 \frac {3x^3}{x^3+1}\ dx} - 3\bigg[x \log x - x\bigg]_0^1 \\ & = {\color{#3D99F6} \log 2 - 3\int_0^1 \left(1 - \frac 1{x^3+1}\right) dx} + 3 \\ & = \log 2 - 3 + 3 \int_0^1 \frac 1{(x+1)(x^2-x+1)} dx + 3 \\ & = \log 2 + 3 \int_0^1 \frac 13 \left(\frac 1{x+1} - \frac {x-2}{x^2-x+1}\right) dx \\ & = \log 2 + \int_0^1 \left(\frac 1{x+1} - \frac {2x-1}{2(x^2-x+1)} + \frac 3{2(x^2-x+1)} \right) dx \\ & = \log 2 + \log (x+1) - \frac {\log (x^2-x+1)} 2 \bigg|_0^1 + \color{#3D99F6} 2 \int_0^1 \frac 1{\frac 43\left(x-\frac 12 \right)^2+1} dx & \small \color{#3D99F6} \text{Let }u = \frac 2{\sqrt 3}\left(x-\frac 12 \right) \\ & = 2\log 2 + \color{#3D99F6} \sqrt 3 \int_{-\frac 1{\sqrt 3}}^{\frac 1{\sqrt 3}} \frac 1{u^2+1} du \\ & = 2\log 2 + \sqrt 3 \tan^{-1} u \ \bigg|_{-\frac 1{\sqrt 3}}^{\frac 1{\sqrt 3}} \\ & = \log 4 + \sqrt 3 \times \frac \pi 3 \\ & = \frac \pi{\sqrt 3} + \log 4 \end{aligned}$

$\implies a+b = 3+4 = \boxed{7}$