Find partial fractions and integrate each term

Write $\dfrac{4+x}{(1-x)(4+x^2)}=\dfrac A{1-x}+\dfrac{Bx+C}{4+x^2}$. Then $A(4+x^2)+(Bx+C)(1-x)=4+x$. Substitution $x=1$ gives $A=1$; comparing x² gives $B=1$, and comparing constants gives $C=0$. Thus the decomposition is $\dfrac1{1-x}+\dfrac x{4+x^2}$.

Integrate the two fractions: $\int\dfrac1{1-x}dx+\int\dfrac x{4+x^2}dx=-\ln|1-x|+\frac12\ln(x^2+4)+C$.