91Ó°ÊÓ

To integrate this rational function, we first use partial fraction decomposition to rewrite the given function as a sum of simpler fractions. The given integrand is \(\frac{1}{8-x^2}\). Factoring the denominator, we have: \begin{align*} 8-x^2 &= (x-\sqrt{8})(x+\sqrt{8}) \end{align*} Using partial fractions: \begin{align*} \frac{1}{8-x^2} &= \frac{A}{x - \sqrt{8}} + \frac{B}{x + \sqrt{8}} \end{align*} Next, let's find the constants A and B.

To find A and B, we need to get rid of the denominators: \begin{align*} 1 &= A(x + \sqrt{8}) + B(x - \sqrt{8}) \end{align*} Now we want two values of x that will help us to determine the unknown constants A and B. By putting x = \(\sqrt{8}\), we have \begin{align*} 1 &= 2\sqrt{8}A \\ A &= \frac{1}{2\sqrt{8}} \end{align*} By putting x = \(-\sqrt{8}\), we have \begin{align*} 1 &= -2\sqrt{8}B \\ B &= \frac{-1}{2\sqrt{8}} \end{align*} Substituting the values of A and B, we get: \begin{align*} \frac{1}{8-x^2} &= \frac{\frac{1}{2\sqrt{8}}}{x-\sqrt{8}} + \frac{\frac{-1}{2\sqrt{8}}}{x+\sqrt{8}} \end{align*} Now, we proceed to integrate.

Now we can integrate term by term: \begin{align*} \int \frac{1}{8-x^2} dx &= \int \left(\frac{\frac{1}{2\sqrt{8}}}{x-\sqrt{8}} + \frac{\frac{-1}{2\sqrt{8}}}{x+\sqrt{8}}\right)dx \\ &= \frac{1}{2\sqrt{8}}\int\frac{1}{x-\sqrt{8}}dx - \frac{1}{2\sqrt{8}}\int\frac{1}{x+\sqrt{8}}dx \end{align*} Both integrals can be solved using a simple u-substitution. Let's integrate the first term: \begin{align*} u &= x-\sqrt{8} \\ du &= dx \\ \int\frac{1}{x-\sqrt{8}}dx &= \int\frac{1}{u}du \end{align*} The integral simplifies to: \begin{align*} \int\frac{1}{u}du &= ln(u)+C_1 \\ &= ln(x-\sqrt{8})+C_1 \end{align*} Now, let's integrate the second term: \begin{align*} v &= x+\sqrt{8} \\ dv &= dx \\ \int\frac{1}{x+\sqrt{8}}dx &= \int\frac{1}{v}dv \end{align*} The integral simplifies to: \begin{align*} \int\frac{1}{v}dv &= ln(v)+C_2 \\ &= ln(x+\sqrt{8})+C_2 \end{align*} Combining the results, we have: \begin{align*} \int \frac{1}{8-x^2} dx &= \frac{1}{2\sqrt{8}}ln(x-\sqrt{8})-\frac{1}{2\sqrt{8}}ln(x+\sqrt{8})+C \\ &= \frac{1}{2\sqrt{8}}\left[ln(x-\sqrt{8})-ln(x+\sqrt{8})\right]+C \end{align*} So, the indefinite integral of \(\frac{1}{8-x^2}\) for \(x > 2\sqrt{2}\) is: \begin{align*} \int \frac{1}{8-x^2} dx = \frac{1}{2\sqrt{8}}\left[ln(x-\sqrt{8})-ln(x+\sqrt{8})\right]+C \end{align*}