# $\int \frac{x^2+4x+4}{(x^2+5x+7)\sqrt{x+2} } dx$

The question is to evaluate $$\int \frac{x^2+4x+4}{(x^2+5x+7)\sqrt{x+2} } dx$$

I tried rewriting the integral as $$\int \frac{dx}{\sqrt{x+2}} - \int\frac{x+3}{(x^2+5x+7)\sqrt{x+2}} dx$$ the first integral is simply $2\sqrt{x+2}$.I tried using Integration by parts on second integral but it got complicated.Any ideas?

• The appearance of $\sqrt{x + 2}$ (and fact that the numerator is $(x + 2)^2$) suggests substituting $u = x - 2$. – Travis Willse Nov 27 '17 at 15:38

With substitution $x+2=u^2$ we have $$\int \frac{x^2+4x+4}{(x^2+5x+7)\sqrt{x+2} } dx=2\int\dfrac{u^4}{u^4+u^2+1}du=2u-2\int\dfrac{u^2+1}{u^4+u^2+1}du$$ for the second write denominator as $$u^4+u^2+1=(u^2+\dfrac12)^2+(\dfrac{\sqrt{3}}{2})^2=\left(u^2+\dfrac{1+i\sqrt{3}}{2}\right)\left(u^2+\dfrac{1-i\sqrt{3}}{2}\right)$$ and use fractions decomposition.