Answer
$\displaystyle{\int_0^1{\frac{3x^2+1}{x^3+x^2+x+1}dx}=\ln 4\sqrt{2}-\frac{\pi}{4}}$
Work Step by Step
$
I=\displaystyle\int_0^1{\frac{3x^2+1}{x^3+x^2+x+1}dx}
$
Say, $f\left( x \right) =x^3+x^2+x+1,$ then $f\left( -1 \right)=0$. Therefore, $\left( x+1 \right)$ is a root of $f\left( x \right)$. Thus, $(x^2+1)$ is the other root.
$I=\displaystyle\int_0^1{\frac{3x^2+1}{\left( x^2+1 \right) \left( x+1 \right)}dx}\\
I=\displaystyle\int_0^1{\frac{3x^2+1+2-2}{\left( x^2+1 \right) \left( x+1 \right)}dx}\\
I=\displaystyle\int_0^1{\frac{3x^2+3-2}{\left( x^2+1 \right) \left( x+1 \right)}dx}\\
I=\displaystyle\int_0^1{\frac{3\left( x^2+1 \right) -2}{\left( x^2+1 \right) \left( x+1 \right)}dx}\\
I=3\displaystyle\int_0^1{\frac{x^2+1}{\left( x^2+1 \right) \left( x+1 \right)}dx}-2\int_0^1{\frac{1}{\left( x^2+1 \right) \left( x+1 \right)}dx}\\
I=3\displaystyle\int_0^1{\frac{1}{x+1}dx}-2\int_0^1{\frac{1}{\left( x^2+1 \right) \left( x+1 \right)}dx}\\$
$\rightarrow 3\displaystyle\int_0^1{\frac{1}{x+1}dx}=
3\left[ \ln \left| x+1 \right| \right] _{0}^{1}$
$\rightarrow -2\displaystyle\int_0^1{\frac{1}{\left( x^2+1 \right) \left( x+1 \right)}dx}\\$
$\displaystyle{\frac{1}{\left( x^2+1 \right) \left( x+1 \right)}\equiv \frac{Ax+B}{x^2+1}+\frac{C}{x+1}\\}
1=\left( Ax+B \right) \left( x+1 \right) +C\left( x^2+1 \right) \\
1=Ax^2+Ax+Bx+B+Cx^2+C$
$x=0\ \therefore \ B+C=1$
$x=1\ \therefore \ 1=A+C+A+B+1 \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0=2A+B+C\ \Rightarrow \ A=-\frac{1}{2}\\
x=2\ \therefore\ 1=4A+4C+2A+2B+1\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0=6A+2C+2C+2B+1\\
\ \ \ \ \ \ \ \ \ \ \ \ -2=6A+2C\\
1=2C\ \Rightarrow \ C=\frac{1}{2}\ \therefore \ B=\frac{1}{2}$
$-2\displaystyle{\int_0^1{\frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1}+\frac{1}{2\left( x+1 \right)}dx}}\\
-2\displaystyle\int_0^1{\frac{1-x}{2\left( x^2+1 \right)}+\frac{1}{2\left( x+1 \right)}dx}\\
-2\displaystyle\int_0^1{-\frac{x-1}{2\left( x^2+1 \right)}+\frac{1}{2\left( x+1 \right)}dx}\\
\displaystyle\int_0^1{\frac{x-1}{x^2+1}dx}-\int_0^1{\frac{1}{x+1}dx}$
$\displaystyle-\int_0^1{\frac{1}{x+1}dx}=-\left[ \ln \left| x+1 \right| \right] _{0}^{1}$
$\displaystyle\int_0^1{\frac{x-1}{x^2+1}dx}\\$
$\begin{matrix}
x=\tan \theta& x^2=\tan ^2\theta\\
\frac{dx}{d\theta}=\sec ^2\theta& dx=\sec ^2\theta \ d\theta\\
\end{matrix}$
$\displaystyle\int_0^{\frac{\pi}{4}}{\frac{\tan \theta -1}{\tan ^2\theta +1}\sec ^2\theta \ d\theta}\\
\displaystyle\int_0^{\frac{\pi}{4}}{\tan \theta -1\ d\theta}=\left[ \ln \left| \sec \theta \right|-\theta \right] _{0}^{\pi /4}$
$I=3\left[ \ln \left| x+1 \right| \right] _{0}^{1}-\left[ \ln \left| x+1 \right| \right] _{0}^{1}+\left[ \ln \left| \sec \theta \right|-\theta \right] _{0}^{\pi /4}\\
I=\displaystyle3\ln 2-\ln 2+\ln \sqrt{2}-\frac{\pi}{4}\\
I=\displaystyle{\ln 4\sqrt{2}-\frac{\pi}{4}}$
