Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int_{0}^{1} \frac{e^{x}}{36-e^{2 x}} d x\) (Give the exact answer and the decimal equivalent. Round to five decimal places.)

Short answer

Exact: \( \frac{1}{12} \ln\left(\frac{5(6-e)}{7(6+e)}\right) \); Decimal: \(-0.09009\).

Step-by-step solution

Apply Substitution

Let \( u = e^x \). Then, \( du = e^x \, dx \) or \( dx = \frac{du}{u} \). For the limits, when \( x = 0 \), \( u = e^0 = 1 \); when \( x = 1 \), \( u = e^1 = e \). Thus, the integral becomes \( \int_{1}^{e} \frac{1}{36-u^2} \, du \).

Express as Partial Fractions

Factor the denominator \( 36-u^2 \) as \( (6-u)(6+u) \). Then express \( \frac{1}{36-u^2} \) as partial fractions: \( \frac{1}{(6-u)(6+u)} = \frac{A}{6-u} + \frac{B}{6+u} \). Solve for \( A \) and \( B \) by equating coefficients: \( 1 = A(6+u) + B(6-u) \), which gives \( A = \frac{1}{12} \) and \( B = -\frac{1}{12} \).

Integrate the Partial Fractions

Substitute \( \frac{1}{12} \) and \( -\frac{1}{12} \) back into the integrals: \[ \frac{1}{12} \int_{1}^{e} \frac{1}{6-u} \, du - \frac{1}{12} \int_{1}^{e} \frac{1}{6+u} \, du. \] These are standard integrals, \( \int \frac{1}{a-x} \, dx = -\ln|a-x| \) and \( \int \frac{1}{a+x} \, dx = \ln|a+x| \). So the integrals become \[ \frac{1}{12} [-\ln|6-u|]_{1}^{e} - \frac{1}{12} [\ln|6+u|]_{1}^{e}. \]

Evaluate the Integrals

Compute each part: \[ \frac{1}{12} ([-\ln|6-e| + \ln|5|]) - \frac{1}{12} ([\ln|6+e| - \ln|7|]). \] Simplify to \[ \frac{1}{12} \ln\left(\frac{5(6-e)}{7(6+e)}\right). \]

Simplify and Approximate

First find the exact value: \[ \frac{1}{12} \ln\left(\frac{5(6-e)}{7(6+e)}\right). \] Calculate using decimals: \( e \approx 2.71828 \), which gives \( 6-e \approx 3.28172 \) and \( 6+e \approx 8.71828 \). Substitute these into our expression to approximate the value: \[ \frac{1}{12} \ln\left(\frac{5 \times 3.28172}{7 \times 8.71828}\right) \approx -0.09009. \]

Key concepts

Substitution in Integrals

When facing an integral of a more complex function, substitution can significantly simplify the process. It involves choosing a new variable, say \( u \), to replace an existing expression within the integral. This process is particularly effective when the derivative of the chosen substitution appears elsewhere in the integral.
For the given integral \( \int_{0}^{1} \frac{e^{x}}{36-e^{2x}} dx \), substitution is used to simplify it to a more manageable form. By setting \( u = e^x \), the differential \( du = e^x \, dx \) allows the equation \( dx = \frac{du}{u} \). You then replace \( e^x \) with \( u \) and rewrite the bounds as \( u = 1 \) when \( x = 0 \) and \( u = e \) when \( x = 1 \).
Thus, the integral becomes \( \int_{1}^{e} \frac{1}{36-u^2} \, du \), turning it into an integral of a rational function, which is easier to evaluate.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions. This technique is beneficial when the integrand can be broken down into simpler terms, making integration more straightforward.
For the integral \( \int_{1}^{e} \frac{1}{36-u^2} \, du \), you start by factoring the denominator: \( 36 - u^2 = (6-u)(6+u) \). The expression \( \frac{1}{(6-u)(6+u)} \) is decomposed into: \( \frac{A}{6-u} + \frac{B}{6+u} \). To find the values of \( A \) and \( B \), set up the equation \( 1 = A(6+u) + B(6-u) \). By solving this, you get \( A = \frac{1}{12} \) and \( B = -\frac{1}{12} \).
This decomposition allows you to split the integral into two simpler integrals, each of which can be integrated using basic logarithmic rules.

Rational Functions

Rational functions are quotients of two polynomials and play a significant role in calculus, particularly in integral calculus. Their integrals often require techniques such as substitution and partial fraction decomposition for evaluation.
In the exercise, the function \( \frac{e^x}{36-e^{2x}} \) is transformed into a rational function through substitution, resulting in \( \frac{1}{36-u^2} \). This formula now represents a ratio of two polynomials, \( 1 \) over \( 36 - u^2 \).
The power of substitution turns transcendental functions into rational functions, which simplifies the process of finding antiderivatives. This conversion often leads to the ability to integrate them by recognizing their structures, such as simple polynomials or products and sums of well-known basic forms.

Natural Logarithm Properties

The integration process often involves natural logarithm properties, especially when dealing with rational functions after substitution and partial fraction decomposition.
From the reduced integral \( \int_{1}^{e} \frac{1}{(6-u)(6+u)} \), you get terms like \( \int \frac{1}{6-u} \, du \) and \( \int \frac{1}{6+u} \, du \), which are standard logarithmic forms. The integral of \( \frac{1}{a-x} \) is \(-\ln|a-x|\) and for \( \frac{1}{a+x} \), it is \(\ln|a+x|\).
These properties simplify the integration to produce logarithmic outputs, where the integral evaluates as \( \frac{1}{12} [-\ln|6-u|] - \frac{1}{12} [\ln|6+u|] \). These forms are key in expressing the solution in terms of the natural logarithm, aiding in both exact evaluation and approximation of these kinds of integrals.