# How do you find the integral intdt/(sqrt(t^2-6t+13)) ?

Sep 10, 2014

By completing the square,
${t}^{2} - 6 t + 13 = \left({t}^{2} - 6 t + 9\right) + 4 = {\left(t - 3\right)}^{2} + {2}^{2}$

So, we can rewrite the integral as
$\int \frac{\mathrm{dt}}{\sqrt{{\left(t - 3\right)}^{2} + {2}^{2}}}$

Let $t - 3 = 2 \tan \theta$.
Rightarrow {dt}/{d theta}=2sec^2 theta Rightarrow dt=2sec^2 theta d theta

by the above substitution,
=int{2sec^2theta d theta}/{sqrt{(2tan theta)^2+2^2)} =int{sec^2theta}/{sqrt{tan^2theta+1}}d theta
by the identity ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$,
$= \int \frac{{\sec}^{2} \theta}{\sqrt{{\sec}^{2} \theta}} d \theta = \int \sec \theta d \theta = \ln | \sec \theta + \tan \theta | + {C}_{1}$
by using
$\tan \theta = \frac{t - 3}{2}$ and $\sec \theta = \sqrt{{\tan}^{2} \theta + 1} = \frac{\sqrt{{t}^{2} - 6 + 13}}{2}$,
we have
$= \ln | \frac{\sqrt{{t}^{2} - 6 + 13}}{2} + \frac{t - 3}{2} | + {C}_{1}$
$= \ln | \frac{\sqrt{{t}^{2} - 6 t + 13} + t - 3}{2} | + {C}_{1}$
by log property,
$= \ln | \sqrt{{t}^{2} - 6 t + 13} + t - 3 | - \ln 2 + {C}_{1}$
by setting $C = \ln 2 + {C}_{1}$,
$= \ln | \sqrt{{t}^{2} - 6 t + 13} + t - 3 | + C$