evaluate $int\frac{dx}{sqrt{x^{2}-6x + 13}}$

Explanation:

Step1: Complete the square in the denominator

First, complete the square for $x^{2}-6x + 13$. We have $x^{2}-6x+13=(x - 3)^{2}-9 + 13=(x - 3)^{2}+4$. So the integral becomes $\int\frac{dx}{\sqrt{(x - 3)^{2}+4}}$.

Step2: Use substitution

Let $u=x - 3$, then $du=dx$. The integral is $\int\frac{du}{\sqrt{u^{2}+4}}$. Now, let $u = 2\tan\theta$, then $du=2\sec^{2}\theta d\theta$. And $\sqrt{u^{2}+4}=\sqrt{4\tan^{2}\theta+4}=2\sec\theta$.

Step3: Simplify the integral

The integral $\int\frac{du}{\sqrt{u^{2}+4}}$ becomes $\int\frac{2\sec^{2}\theta d\theta}{2\sec\theta}=\int\sec\theta d\theta$.

Step4: Integrate $\sec\theta$

The integral of $\sec\theta$ is $\ln|\sec\theta+\tan\theta|+C$. Since $u = 2\tan\theta$, then $\tan\theta=\frac{u}{2}$ and $\sec\theta=\frac{\sqrt{u^{2}+4}}{2}$.

Step5: Back - substitute

Substitute back $u=x - 3$. We get $\ln|\frac{\sqrt{(x - 3)^{2}+4}}{2}+\frac{x - 3}{2}|+C=\ln|\sqrt{(x - 3)^{2}+4}+(x - 3)|-\ln2 + C$. Since $-\ln2+C$ is also a constant, the final result is $\ln|x - 3+\sqrt{x^{2}-6x + 13}|+C$.

Answer:

$\ln|x - 3+\sqrt{x^{2}-6x + 13}|+C$