would the three answers of this integral $\int \frac{1}{a x^2+b x+c} d x$ make a continuous function

Question

so this integral $$ \int \frac{1}{a x^2+b x+c} d x $$ has three answers depending on the discriminant $$b^2-4 a c$$ whether is positive zero or negative

1. $b^2-4 a c=0$ $$ \frac{-2}{2 a x+b} $$
2. $b^2-4 a c>0$ $$ \frac{1}{\sqrt{b^2-4 a c}} \ln \left|\frac{2 a x+b-\sqrt{b^2-4 a c}}{2 a x+b+\sqrt{b^2-4 a c}}\right| $$
3. $b^2-4 a c<0$ $$ \frac{2}{\sqrt{4 a c-b^2}} \tan ^{-1}\left(\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right) $$

so would a function like $$ \begin{aligned} &-\frac{2}{2 a x+b}\left\{b^2-4 a c=0\right\}\\ &\frac{\ln \left(\left|\frac{2 a x+b-\sqrt{b^2-4 a c}}{2 a x+b+\sqrt{b^2-4 a c}}\right|\right)}{\sqrt{b^2-4 a c}}\left\{b^2-4 a c>0\right\}\\ &\frac{2 \tan ^{-1}\left(\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right)}{\sqrt{4 a c-b^2}}\left\{b^2-4 a c<0\right\} \end{aligned} $$ be continuous