Question

$\int \sqrt{x^2+4x+6}dx$ is equal to
(where $C$ is integration constant)

• A. $\frac{x+2}{2}\sqrt{x^2+4x+6}+\ln |x+2+\sqrt{x^2+4x+6}|+C$
• B. $\ln |x+2+\sqrt{x^2+4x+6}|+C$
• C. $\frac{x}{2}\sqrt{x^2+4x+6}+\ln |x+\sqrt{x^2+4x+6}|+C$
• D. $\frac{1}{2}{\tan }^{-1}\left(\frac{x+2}{\sqrt{2}}\right)+C$

Answer 1

The correct option is A $\frac{x+2}{2}\sqrt{x^2+4x+6}+\ln |x+2+\sqrt{x^2+4x+6}|+C$

Let
$I=\int \sqrt{x^2+4x+6}dx=\int \sqrt{(x+2{)}^2+(\sqrt{2}{)}^2}dx\Rightarrow I=\frac{(x+2)}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\ln \left|\right(x+2)+\sqrt{x^2+4x+6}|+C[\because \int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln |x+\sqrt{x^2+a^2}|+C]\therefore I=\frac{(x+2)}{2}\sqrt{x^2+4x+6}+\ln \left|\right(x+2)+\sqrt{x^2+4x+6}|+C$