Question

Determine the following indefinite integrals. \(\int \frac{d x}{8-x^{2}}, x>2 \sqrt{2}\)

Short answer

Answer: \(\int \frac{1}{8-x^2} dx = \frac{1}{2\sqrt{8}}\left[ln(x-\sqrt{8})-ln(x+\sqrt{8})\right]+C\)

Step-by-step solution

Partial fraction decomposition

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.

Determine 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.

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*}

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a powerful tool used to simplify complex rational expressions. This technique involves breaking down a fraction into a sum of simpler fractions, making integration easier.
In our problem, the integrand is \( \frac{1}{8-x^2} \), which poses integration challenges due to its polynomial in the denominator. First, we factor the denominator as \((x-\sqrt{8})(x+\sqrt{8})\).

• The idea is to express \( \frac{1}{8-x^2} \) as \( \frac{A}{x - \sqrt{8}} + \frac{B}{x + \sqrt{8}} \)
• By clearing denominators, we simplify this equation to \(1 = A(x+\sqrt{8}) + B(x-\sqrt{8})\)

Choosing strategic values for \(x\) simplifies finding \(A\) and \(B\):

• Substitute \( x = \sqrt{8} \) to find \( A = \frac{1}{2\sqrt{8}} \)
• Substitute \( x = -\sqrt{8} \) to find \( B = \frac{-1}{2\sqrt{8}} \)

Substituting these constants back into the partial fractions gives us separate, integrable terms and marks the completion of this method. This allows us to transform an otherwise complex integration into a manageable process.

U-Substitution

Integration by u-substitution involves changing variables to simplify the integration process. When dealing with logarithmic and trigonometric functions, u-substitution can be a lifesaver.
In our exercise, after breaking down the original fraction, we apply u-substitution to each term.

• For the term \( \frac{1}{x-\sqrt{8}} \), let \( u = x-\sqrt{8} \), thus \( du = dx \)
• For the term \( \frac{1}{x+\sqrt{8}} \), let \( v = x+\sqrt{8} \), thus \( dv = dx \)

This substitution converts both integrals into the simpler form \( \int \frac{1}{u} du \) and \( \int \frac{1}{v} dv \).
This streamlines the integration to a natural logarithm form, \( \ln|u| + C \), providing the integrated terms efficiently. Through u-substitution, we have essentially re-shaped the problem into something accessible, showcasing the beauty of mathematical transformation.

Logarithmic Integration

Logarithmic integration appears when the integrand simplifies to \( \frac{1}{u} \). It's a fundamental type of integral that transforms into a natural logarithm, \( \ln|u| + C \).
This type of integration is prevalent in the given problem as both decomposed fractions simplify to this form.

• The integral \( \int \frac{1}{x-\sqrt{8}} dx \) becomes \( \ln|x-\sqrt{8}| + C_1 \)
• Likewise, \( \int \frac{1}{x+\sqrt{8}} dx \) becomes \( \ln|x+\sqrt{8}| + C_2 \)

These separate integrals combine to express the result through logarithmic properties:
\[\ln(x-\sqrt{8}) - \ln(x+\sqrt{8})\]
By integrating in this manner, we take advantage of the simplicity of logarithms. The indecomposable parts, represented as natural logarithms, express the original complex function in a succinct integrated form. Thus, logarithmic integration provides a neatly packaged result, highlighting its efficient role in solving integral problems.