1.16 Combining terms

Consider the special case $b=0$, $r=1$ in case (ii). Since the denominator is irreducible, it follows that $c>0$ and so there exists $\alpha>0$ such that $c=\alpha^{2}$. In this case we integrate $\frac{Hx+K}{x^{2}+\alpha^{2}}$ as follows: we split it as

and integrate each term separately.

a) For the first term we make the substitution $u=x^{2}+\alpha^{2}$. Then $\frac{du}{dx}=2x$, so $2x\,dx=du$ and therefore:

which equals $\frac{H}{2}\log(x^{2}+\alpha^{2})+C$.