Question Detail

The denominator ( x^2+4x+7 ) can be completed to a square:( x^2+4x+7 = (x^2+4x+4) + 3 = (x+2)^2 + (\sqrt{3})^2 )The integral becomes ( \int \frac{dx}{(x+2)^2 + (\sqrt{3})^2} ).This is in the form of ( \int \frac{du}{u^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C ).Here, ( u = x+2 ) and ( a = \sqrt{3} ).So the integral evaluates to ( \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x+2}{\sqrt{3}}\right) + C ).