Integral Integral5

$\begin{aligned} I & = \int_0^1 \log^2\left(\frac 1x + 1\right) dx = \int_0^1 \log^2\left(\frac {x+1}x \right) dx = \int_0^1 \left(\log(x+1)-\log x \right)^2 dx \\ & = \int_0^1 \left(\log^2(x+1)- 2 \log x \log(x+1) + \log^2 x \right) dx \\ & = \int_0^1 \log^2(x+1)\ dx - 2 \int_0^1 \log x \log(x+1)\ dx + \int_0^1 \log^2 x \ dx \\ & = 2\log^2 2 - 4\log 2 + 2 + 4\log 2 - 4 + \frac {\pi^2}6 + 2 \\ & = \frac {\pi^2}6 + 2\log^2 2 \end{aligned}$

Therefore, $a+b+c+d = 2+6+2+2 = \boxed{12}$ .

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Notes:

$\begin{aligned} I_1 & = \int_0^1 \log^2 (x+1) \ dx & \small \color{#3D99F6} \text{Let }u = x+1 \implies du = dx \\ & = \int_1^2 \log^2 u \ du & \small \color{#3D99F6} \text{By integration by parts} \\ & = \log u (u\log u - u)\bigg|_1^2 - \int_1^2 (\log u - 1)\ du \\ & = \log 2 (2\log 2 - 2) - \bigg[u\log u - 2u \bigg]_1^2 \\ & = 2\log^2 2 - 2\log 2 - (2\log 2 - 4 + 2) \\ & = 2\log^2 2 - 4\log 2 + 2 \end{aligned}$

$\begin{aligned} I_2 & = \int_0^1 \log x \log (x+1) \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = \log x ((x+1)\log (x+1)-x)\bigg|_0^1 - \int_0^1 \frac {(x+1)\log (x+1)-x}x dx \\ & = 0 - \int_0^1 \log (x+1) \ dx - {\color{#3D99F6} \int_0^1 \frac {\log (x+1)}x dx} + \int_0^1 dx \\ & = - \bigg[(x+1)\log (x+1) - x\bigg]_0^1 - {\color{#3D99F6} \int_0^1 \frac 1x \sum_{n=1}^\infty \frac {(-1)^{n+1}x^n}n dx} +1 \\ & = - 2\log 2 + 1 - {\color{#3D99F6} \int_0^1 \sum_{n=1}^\infty \frac {(-1)^{n+1}x^{n-1}}n dx} +1 \\ & = - 2\log 2 + 2 - {\color{#3D99F6} \sum_{n=1}^\infty \frac {(-1)^{n+1}x^n}{n^2}\bigg|_0^1} \\ & = - 2\log 2 + 2 - \left(\frac 1{1^2}-\frac 1{2^2}+\frac 1{3^2} - \frac 1{4^2}+\frac 1{5^2} - \cdots \right) \\ & = - 2\log 2 + 2 - \left(\frac 1{1^2}+\frac 1{2^2} +\frac 1{3^2}+ \cdots - 2\left(\frac 1{2^2}+\frac 1{4^2} +\frac 1{6^2}+ \cdots \right) \right) \\ & = - 2\log 2 + 2 - \left({\color{#3D99F6}\frac 1{1^2}+\frac 1{2^2} +\frac 1{3^2}+ \cdots} - \frac 2{2^2} \left({\color{#3D99F6}\frac 1{1^2}+\frac 1{2^2} +\frac 1{3^2}+ \cdots} \right) \right) & \small \color{#3D99F6} \text{Riemann's zeta function }\zeta (n) = \sum_{k=1}^\infty \frac 1{k^n} \\ & = - 2\log 2 + 2 - \left({\color{#3D99F6}\zeta (2)} - \frac 12 \left({\color{#3D99F6}\zeta (2)} \right) \right) & \small \color{#3D99F6} \text{and }\zeta (2) = \sum_{k=1}^\infty \frac 1{k^2} = \frac {\pi^2} 6 \\ & = - 2\log 2 + 2 - \frac {\pi^2}{12} \end{aligned}$

$\begin{aligned} I_3 & = \int_0^1 \log^2 x \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = \log x (x\log x - x)\bigg|_0^1 - \int_0^1 (\log x -1) \ dx \\ & = 0 - x\log x +2 x \ \bigg|_0^1 \\ & = 2 \end{aligned}$