By expressing the following in partial fractions evaluate the given integral. Remember to select the correct form for the partial fractions. $$ \int \frac{1}{x^{2}-2 x-1} \mathrm{~d} x $$

The denominator of the given integrand is a quadratic, given by: $$x^2 - 2x - 1$$ Unfortunately, factoring this quadratic does not yield rational factors, and as such, we cannot use the method of partial fractions in its current form directly. However, we can complete the square to get a more suitable form for integration. Let us complete the square for the quadratic expression: - Move constant term to the right side: $$x^2 - 2x = 1$$ - Calculate the square of half the coefficient of the linear term. In this case, \(\frac{-2}{2} = -1\), hence \((-1)^2 = 1\), and add this square to both sides: $$x^2 - 2x + 1 = 1 + 1 \Rightarrow (x-1)^2 = 2$$ Now, the denominator can be expressed as: $$(x-1)^2 - 2$$

Now that we have the quadratic written in a more suitable form, we can notice that this denominator can be compared to the term \((x-a)^2 + b^2\). In our case, \(a=1\) and \(b=\sqrt{2}\), and the form cannot be further decomposed. Therefore, we can rewrite the integral as: $$ \bigintsss \frac{1}{(x-1)^2 - (\sqrt{2})^2} dx $$

Upon inspection, one can recognize that the integrand is in the form of inverse tangent function derivative, implying that the arctangent substitution should be used. To do this, we make a substitution: $$ x - 1 = \sqrt{2} \tan{\theta} $$ resulting in: $$ dx = \sqrt{2} \sec^2{\theta} d\theta $$ The integral becomes: $$ \bigintsss \frac{\sqrt{2} \sec^2{\theta} d\theta}{2\tan^2{\theta}} = \frac{\sqrt{2}}{2} \bigintsss \frac{\sec^2{\theta}}{\tan^2{\theta}} d\theta $$

We can simplify the integrand as follows: $$ \frac{\sqrt{2}}{2} \bigintsss \frac{\sec^2{\theta}}{\tan^2{\theta}} d\theta = \frac{\sqrt{2}}{2} \bigintsss \frac{1}{\sin^2{\theta} \cos^2{\theta}} d\theta = \frac{\sqrt{2}}{2} \bigintsss \frac{1}{(1-\cos^2{\theta}) \cos^2{\theta}} d\theta $$ We now make the following substitution: $$ u = \cos{\theta}, du = -\sin{\theta} d\theta $$ We get: $$ \frac{\sqrt{2}}{2} \bigintsss \frac{1}{u^2(1-u^2)}(-du) = -\frac{\sqrt{2}}{2} \bigintsss \frac{1}{u^2(1-u^2)} du $$ We now apply partial fraction decomposition to the integrand: $$ -\frac{\sqrt{2}}{2} \bigintsss \frac{1}{u^2(1-u^2)} du = -\frac{\sqrt{2}}{4}\left( \bigintsss \frac{1}{u^2} du - \bigintsss \frac{1}{1-u^2} du \right) $$ Integrating the two expressions separately, we get: $$ -\frac{\sqrt{2}}{4} \left( -\frac{1}{u} - \frac{1}{2} \ln\left|\frac{1+u}{1-u}\right| \right) = -\frac{\sqrt{2}}{4}\left( -\frac{1}{\cos{\theta}} - \frac{1}{2} \ln\left|\frac{1+\cos{\theta}}{1-\cos{\theta}}\right|\right) $$ Now, substitute back the original expression for \(u = \cos{\theta}\): $$ -\frac{\sqrt{2}}{4}\left( -\frac{1}{\sqrt{2}\tan{\theta}} - \frac{1}{2} \ln{\left|\frac{1+ \sqrt{2}\tan{\theta}}{1-\sqrt{2}\tan{\theta}}\right|} \right) $$ Finally, substitute back the original expression for \(\tan{\theta}\): $$ -\frac{\sqrt{2}}{4}\left( -\frac{1}{\sqrt{2}(x-1)} - \frac{1}{2} \ln{\left|\frac{1+ \sqrt{2}(x-1)}{1-\sqrt{2}(x-1)}\right|} \right) $$ The final integral is therefore: $$ \int \frac{1}{x^{2}-2 x-1} \mathrm{~d} x = -\frac{\sqrt{2}}{4}\left( -\frac{1}{\sqrt{2}(x-1)} - \frac{1}{2} \ln{\left|\frac{1+ \sqrt{2}(x-1)}{1-\sqrt{2}(x-1)}\right|}\right) + C $$ where \(C\) is the integration constant.