Question Detail

To evaluate this integral, we first complete the square in the denominator:$x^2 + 4x + 7 = (x^2 + 4x + 4) + 3 = (x+2)^2 + (\sqrt{3})^2$Now the integral becomes:$\int \frac{dx}{(x+2)^2 + (\sqrt{3})^2}$This integral is in the standard form $\int \frac{du}{u^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$.Here, let $u = x+2$, so $du = dx$. And $a = \sqrt{3}$.Substituting these values, we get:$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x+2}{\sqrt{3}}\right) + C$