$\begin{aligned}
I&=\displaystyle\int\limits_{0}^{1}\left(3x^2+\mathrm{e}^x+\dfrac{1}{x+1}\right)\mathrm{d}x\\
&=\left(x^3+\mathrm{e}^x+\ln|x+1|\right)\bigg|_0^1\\
&=\left(1^3+\mathrm{e}^1+\ln|1+1|\right)-\left(0^3+\mathrm{e}^0+\ln|0+1|\right)\\
&=\mathrm{e}+\ln2.
\end{aligned}$