Integral Integral16

$\begin{aligned} I & = \int_0^1 \log \left(\sum_{k=0}^5 x^k \right) dx \\ & = \int_0^1 \log \left(\frac {1-x^6}{1-x} \right) dx \\ & ={\color{#3D99F6}\int_0^1 \log \left(1-x^6\right) dx} -\color{#D61F06} \int_0^1 \log \left(1-x\right) dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = {\color{#3D99F6}x\log (1-x^6)\bigg|_0^1 + \int_0^1 \frac {6x^6}{1-x^6} dx} -\color{#D61F06} \int_0^1 \log x\ dx & \small \color{#D61F06} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = {\color{#3D99F6}x\log (1-x^6)\bigg|_0^1 - 6 \int_0^1 \left(\frac {x^6}{x^6-1}\right) dx} -\color{#D61F06} x \log x - x \bigg|_0^1 \\ & = x\log (1-x^6)\bigg|_0^1 - 6 \int_0^1 \left({\color{#3D99F6}\frac 1{x^6-1}}+1\right) dx -\color{#D61F06} (- 1)\end{aligned}$

$\begin{aligned} \ \ & = x\log (1-x^6) - 6 \int_0^1 {\color{#3D99F6}\frac 1{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}} dx - 6 \int_0^1 dx + 1 \\ & = x\log (1-x^6)\bigg|_0^1 - 6 \int_0^1 {\color{#3D99F6}\frac 16 \left( \frac 1{x-1} - \frac 1{x+1} + \frac {x-2}{x^2-x+1} - \frac {x+2}{x^2+x+1} \right)} dx - 5 \\ & = x\log (1-x^6)\bigg|_0^1 - \log(|x-1|)\bigg|_0^1 + \log(x+1)\bigg|_0^1 - \frac 12 \int_0^1 \frac {2x-1-3}{x^2-x+1}dx + \frac 12 \int_0^1 \frac {2x+1+3}{x^2+x+1} dx - 5 \\ & = x\log (1-x^6)\bigg|_0^1 - \log(1-x)\bigg|_0^1 + \log 2 + \frac 12 \log \left(\frac {x^2+x+1}{x^2-x+1}\right)\bigg|_0^1 + 2 \int_0^1 \frac 1{\frac 2{\sqrt 3}\left(x-\frac 12\right)+1}dx + 2 \int_0^1 \frac 1{\frac 2{\sqrt 3}\left(x-\frac 12\right)+1}dx - 5 \\ & = x\log \left(\frac {1-x^6}{1-x} \right) \bigg|_0^1 + \log 2 + \frac {\log 3}2 + \sqrt 3 \tan^{-1} \left(\frac {2x-1}{\sqrt 3}\right) \bigg|_0^1 + \sqrt 3 \tan^{-1} \left(\frac {2x+1}{\sqrt 3}\right) \bigg|_0^1 - 5 \\ & = x\log \left(\sum_{k=0}^5 x^k \right) \bigg|_0^1 + \log 2 + \frac {\log 3}2 + \frac {\pi\sqrt 3}6 + \frac {\pi\sqrt 3}3 - 5 \\ & = \log 6 + \log 2 + \frac {\log 3}2 + \frac {\pi \sqrt 3}2 - 5 \\ & = \frac 12 (-10 + \pi \sqrt 3 + \log 432 ) \end{aligned}$

$\implies a+b+c = 10+3+432 = \boxed{445}$