Evaluate $\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$ Any idea on how to solve the following definite integral?

$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$

I have tried to parameterize the integral like $\ln{(a^2x^2+1)}$ or $\ln{(x^2+a^2)}$, which don't seem to work.

\begin{equation}
 I = \int_0^1 \frac{\ln\left(x^2 + 1\right)}{x + 1}\:dx
\end{equation}
Here let:
\begin{equation}
 I(t) = \int_0^1 \frac{\ln\left(tx^2 + 1\right)}{x + 1}\:dx
\end{equation}
We observe that $I = I(1)$ and $I(0) = 0$. Thus, 
\begin{align}
 I'(t) &= \int_0^1 \frac{x^2}{\left(tx^2 + 1\right)\left(x + 1\right)}\:dx = \frac{1}{t + 1}\int_0^1 \left[\frac{1}{x + 1} + \frac{x}{tx^2 + 1} - \frac{1}{tx^2 + 1} \right]\:dx \\
&= \frac{1}{t + 1}\bigg[\ln(x + 1) + \frac{1}{2t}\ln\left(tx^2 + 1 \right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}x\right) \bigg]_0^1 \\
&=\frac{1}{t + 1} \bigg[\ln(2) + \frac{1}{2t}\ln\left(t + 1\right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}\right)\bigg]
\end{align}
Thus, 
\begin{align}
I(t) &= \int\frac{1}{t + 1} \bigg[\ln(2) + \frac{1}{2t}\ln\left(t + 1\right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}\right)\bigg]\:dt \\
&=  \int\frac{\ln(2)}{t + 1}\:dt + \int \frac{1}{2t\left(t + 1\right)}\ln\left(t + 1\right)\:dt + \int  \frac{1}{\sqrt{t}\left(t + 1\right)}\arctan\left(\sqrt{t}\right)\:dt \\
&= I_1 + I_2 + I_3 
\end{align}
Solve each individually:
For $I_1$:
\begin{equation}
I_1 =  \int\frac{\ln(2)}{t + 1}\:dt = \ln(2)\ln(t + 1)
\end{equation}
For $I_2$:
\begin{align}
 I_2 &= \int\frac{1}{2t\left(t + 1\right)}\ln\left(t + 1\right)\:dt = \frac{1}{2}\left[ \int \left(\frac{1}{t} - \frac{1}{t + 1} \right)\ln(t + 1)\:dt\right]\\
& =\frac{1}{2} \left[\int \frac{\ln(t + 1)}{t}\:dt - \int \frac{\ln(t + 1)}{t + 1}\:dt \right] \\
&= \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right]
\end{align}
Where $\operatorname{Li}_{2}(t)$ is Dilogarithm function. 
For $I_3$:
\begin{equation}
 \int  \frac{1}{\sqrt{t}\left(t + 1\right)}\arctan\left(\sqrt{t}\right)\:dt = \arctan^2\left(\sqrt{t}\right)
\end{equation}
Combining we arrive at:
\begin{equation}
 I(t) = \ln(2)\ln(t + 1) +  \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right] + \arctan^2\left(\sqrt{t}\right) + C
\end{equation}
Where $C$ is the constant of integration. 
Using $I(0) = 0$ we see that $C = 0$. Thus
\begin{equation}
 I(t) = \ln(2)\ln(t + 1) +  \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right] + \arctan^2\left(\sqrt{t}\right) 
\end{equation}
And finally:
\begin{align}
 I = I(1) &= \ln(2)\ln(1 + 1) +  \frac{1}{2} \left[-\operatorname{Li}_{2}(-1) - \frac{1}{2}\ln^2(1 + 1) \right] + \arctan^2\left(\sqrt{1}\right) \\
&=\ln^2(2) +  \frac{1}{2} \left[-\frac{\pi^2}{12} - \frac{1}{2}\ln^2(2)\right] + \frac{\pi^2}{16} \\
&= \frac{3}{4}\ln^2(2) - \frac{\pi^2}{48}
\end{align}
