Question Detail

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