Ta có:
$\int\limits_0^1\dfrac{(x+1)^2}{x^2+1}dx$
$=\int\limits_0^1\left(1+\dfrac{2x}{x^2+1}\right)dx$
$=\int\limits_0^1dx+\int\limits_0^1\dfrac{d(x^2+1)}{x^2+1}$
$=x\left|\begin{array}{l}1\\0\end{array}\right.+\ln(x^2+1)\left|\begin{array}{l}1\\0\end{array}\right.$
$=1+\ln2$