Compute $\int_3^6 \frac{\sqrt{3}}{x^2 – 6x +12} \,dx .$
I know this is a simple(?) integration problem, but I'm really having a hard time trying to solve it. $\int_3^6 \frac{\sqrt{3}}{x^2 – 6x +12} \,dx = \int_3^6 \frac{\sqrt{3}}{(x-3)^2+3} \,dx$, and I feel like a trig substitution would help, but letting $x-3 = \tan \theta$ left me stuck, and in another case where $x-3 =\sec \theta$, I got to (ignoring limits for now) $\sqrt{3} \int \sin \theta \cdot \frac{\sec^2 \theta}{\tan^2 \theta + 4} \, d\theta$, and I feel like that could get me somewhere, but I am stuck again.
Best Answer
Instead of the trigonometric substitution, substituting $u = \frac1{\sqrt 3}\left(x-3\right) \implies du = \frac{1}{\sqrt 3}\, dx$, gives us: $$I = \frac13 \int_{3}^{6} \frac{\sqrt 3}{\left(\color{red}{\frac{x-3}{\sqrt 3}}\right)^2+1}\, dx $$ $$= \int_{0}^{\sqrt 3} \frac{1}{u^2+1}\, du$$ which is a standard result.