91Ó°ÊÓ

The first step is to factor the denominator of the given expression. The denominator is a quadratic of the form \(x^2 - 1\), which can be factored as \((x - 1)(x + 1)\). Now, we can write the expression as: \(\frac{1}{(x - 1)(x + 1)}\).

The next step is to decompose the expression into partial fractions. We want to express \(\frac{1}{(x - 1)(x + 1)}\) as: $$\frac{1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$$

Multiply both sides by \((x - 1)(x + 1)\) to get rid of the denominators: $$1 = A(x + 1) + B(x - 1)$$ Now, solve for the constants \(A\) and \(B\). To find \(A\), set \(x = 1\): $$1 = A(1 + 1)$$ $$A = \frac{1}{2}$$ To find \(B\), set \(x = -1\): $$1 = B(-1 - 1)$$ $$B = -\frac{1}{2}$$ So, our expression becomes: $$\frac{1}{(x - 1)(x + 1)} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1}$$

Integrate the decomposed expression: $$\int\frac{1}{x^2 - 1}dx = \int\left(\frac{1/2}{x - 1} - \frac{1/2}{x + 1}\right)dx = \frac{1}{2}\ln|x - 1| - \frac{1}{2}\ln|x + 1| + C$$ So, the integral evaluated using partial fractions is: $$\frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right| + C$$ #Method 2: Trigonometric Substitution#

Let \(x = \cosh{u}\). Thus, \(dx = \sinh{u}du\). Our integral becomes: $$\int\frac{1}{x^2 - 1}dx = \int\frac{1}{(\cosh^2{u} - 1)}\sinh{u}du$$

Recall the identity \(\cosh^2u - 1 = \sinh^2u\). Substitute this into the expression: $$\int\frac{1}{(\cosh^2{u} - 1)}\sinh{u}du = \int\frac{1}{\sinh^2u}\sinh{u}du = \int\frac{1}{\sinh{u}}du$$

Now, integrate the simplified expression: $$\int\frac{1}{\sinh{u}}du = \int\frac{\cosh{u}}{\sinh{u}\cosh{u}}du = \int\frac{\cosh{u}}{\sinh^2{u} + \cosh^2{u} - \cosh^2{u}}du = \int\text{csch}{u}du = \ln\left|\tanh\left(\frac{u}{2}\right)\right| + C$$

Now, substitute back \(u\) with \(x\): $$\ln\left|\tanh\left(\frac{u}{2}\right)\right| = \ln\left|\tanh\left(\frac{\cosh^{-1}{x}}{2}\right)\right| = \ln\left|\frac{\exp(\cosh^{-1}{x}/2) - \exp(-\cosh^{-1}{x}/2)}{\exp(\cosh^{-1}{x}/2) + \exp(-\cosh^{-1}{x}/2)}\right|$$ Recall the definitions of \(\cosh\) and \(\sinh\). This expression simplifies to: $$\ln\left|\frac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}}\right| = \ln\left|\frac{x - 1}{x + 1}\right|$$ So, the integral evaluated using trigonometric substitution is: $$\ln\left|\frac{x - 1}{x + 1}\right| + C$$ #Reconcile the two answers# Both methods give the same expression for the integral: Method 1: $$\frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right| + C$$ Method 2: $$\ln\left|\frac{x - 1}{x + 1}\right| + C$$ Considering a constant can be any real value, we can see that both expressions are equivalent. This confirms that both methods yield the same result for the integral of \(\frac{1}{x^2 - 1}\) for \(x > 1\).