Statement – I: If a>0 and b²-4ac<0 then the value of the integral $\int \frac{dx}{a{x}^2+bx+c}$ will be the type $\mu {\tan }^{-1}\frac{x+A}{B}+C$, where A,B,C, $\mu$  are constants.

Statement – II: a>0 and b²-4ac<0, then ax²+bx+c can be written as the sum of two squares.

$$\begin{gathered} If a>0 and b^2-4ac<0 then \\ \int \frac{1}{a{x}^2+bx+c}dx=\int \frac{1}{a(x^2+2({\displaystyle \frac{b}{2a}})x+{\displaystyle \frac{b^2}{4a^2}}-{\displaystyle \frac{b^2}{4a^2}}+{\displaystyle \frac{c}{a}})}dx \\ =\int \frac{1}{a{(x+{\displaystyle \frac{b}{2a}})}^2+{\displaystyle \frac{4ac-b^2}{4a}}}dx \\ =\int \frac{1}{a{(x+\frac{b}{2a})}^2+k^2}dx where k^2=\frac{4ac-b^2}{4a}>0 \\ =\frac{1}{a}\frac{1}{{\displaystyle \frac{k}{\sqrt{a}}}}{\tan }^{-1}\left(\frac{x+\frac{b}{2a}}{{\displaystyle \frac{k}{\sqrt{a}}}}\right)+C \\ =\mu {\tan }^{-1}\left(\frac{x+A}{B}\right)+C where A,B,C,\mu are cons\tan ts \end{gathered}$$

Assertion is correct, Reason is correct and Reason is the correct explanation of Assertion