Find: ∫ (x^2)/(x^2+6 x+12) d x

Let $\displaystyle I = \int \dfrac{x^2}{x^2 + 6x + 12}\, dx$\[= \int \left[1 - \frac{6x + 12}{x^2 + 6x + 12}\right] dx\]\[\int \left[1 - \frac{6(x + 2)}{x^2 + 6x + 12}\right] dx\]\[x - 6 \int \frac{x + 2}{x^2 + 6x + 12}\, dx \quad \text{------------------ (1)}\]Let $\displaystyle I_1 = \int \dfrac{x + 2}{x^2 + 6x + 12}\, dx$\[x + 2 = A \frac{d}{dx}\left(x^2 + 6x + 12\right) + B\]\[\Rightarrow x + 2 = A(2x + 6) + B\]\[\Rightarrow A = \frac{1}{2} \text{ and } B = -1\]So, $\displaystyle I_1 = \dfrac{1}{2} \int \dfrac{2x + 6}{x^2 + 6x + 12}\, dx - \int \dfrac{1}{x^2 + 6x + 12}\, dx$\[= \frac{1}{2} \log\left|x^2 + 6x + 12\right| - \int \frac{1}{(x + 3)^2 + (\sqrt{3})^2}\, dx\]\[= \frac{1}{2} \log\left|x^2 + 6x + 12\right| - \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x + 3}{\sqrt{3}}\right) + C\]From ($\displaystyle 1$), we have\[I = x - 3 \log\left|x^2 + 6x + 12\right| + 2\sqrt{3} \tan^{-1}\left(\frac{x + 3}{\sqrt{3}}\right) + C\]