Decompose a linear and irreducible quadratic denominator

Clear denominators: $x^2+x=A(x^2+1)+(Bx+C)(x-1)$. At $x=1$, $2=2A$ so $A=1$; at $x=0$, $0=1-C$ so $C=1$; at $x=2$, $6=5+(2B+1)$ so $B=0$. Thus the decomposition is $1/(x-1)+1/(x^2+1)$.

$\int\frac{x^2+x}{(x-1)(x^2+1)}\,dx=\int\bigl(\frac1{x-1}+\frac1{x^2+1}\bigr)\,dx=\ln|x-1|+\tan^{-1}x+C$, for $x\ne1$. The Teacher identifies the logarithmic-derivative form and the arctangent form separately.