Question

Assertion :If $a>0$ and $b^2-4ac<0.$ then the value of the integral $\int \frac{dx}{a{x}^2+bx+c}$ will be of the type $\mu {\tan }^{-1}\left(\frac{x+A}{B}\right)+C;$ where $A,B,C,\mu$ are constant. Reason: If $a>0,b^2-4ac<0,$ then $a{x}^2+bx+c$ can be written as sum of two squares.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

If $a>0$ and $b^2-4ac<0,$ then

$a{x}^2+bx+c=a{(x+\frac{b}{2a})}^2+\frac{4ac-b^2}{4a}$

$\Rightarrow \int \frac{dx}{a{x}^2+bx+c}=\int \frac{dx}{a(x+\frac{b}{2a})+k^2}$

where $k^2=\frac{4ac-b^2}{4a}>0$

which will of the type $\frac{1}{a}.\frac{1}{\frac{k}{\sqrt{a}}}{\tan }^{-1}\frac{x+\frac{b}{2a}}{\frac{k}{\sqrt{a}}}+C=\mu {\tan }^{-1}\left(\frac{x+A}{B}\right)+C$

where $\mu =\frac{1}{\sqrt{a}k},A=\frac{b}{2a},B=\frac{k}{\sqrt{a}}$ are constants