91Ó°ÊÓ

Calculate the integral below by partial fractions and by using the indicated substitution. Be sure that you can show how the results you obtain are the same. $$ \int \frac{2 x}{x^{2}-25} d x $$ First, rewrite this with partial fractions: \(\int \frac{2 x}{x^{2}-25} d x=\int\) _____________ \(d x+\int\) _______________ \(d x=\) ______________ + ______________ \(+C\) (Note that you should not include the +C in your entered answer, as it has been provided at the end of the expression.) Next, use the substitution \(w=x^{2}-25\) to find the integral: \(\int \frac{2 x}{x^{2}-25} d x=\int\) ______________ \(d w=\) ________________ \(+C=\) _______________ \(+C\) (For the second answer blank, give your antiderivative in terms of the variable \(w .\) Again, note that you should not include the +C in your answer.)

Step by step solution

01

- Set up partial fractions

The integral given is \ \( \int \frac{2 x}{x^{2}-25} d x \) \ First, recognize that \( x^{2} - 25 \) can be factored as \( (x - 5)(x + 5) \). Therefore, the integrand can be decomposed into partial fractions: \ \( \frac{2x}{(x - 5)(x + 5)} \)

02

- Partial Fraction Decomposition

Using partial fraction decomposition, \( \frac{2x}{(x - 5)(x + 5)} \) can be written as \( \frac{A}{x-5} + \frac{B}{x+5} \). To find the constants \( A \) and \( B \), set up the equation: \ \( 2x = A(x+5) + B(x-5) \)

03

- Solve for constants A and B

Expanding and equating coefficients of like terms, we get: \ \( 2x = Ax + 5A + Bx - 5B \) \ Grouping by powers of \( x \): \ \( 2x = (A + B)x + (5A - 5B) \) \ This gives two equations: \ \( A + B = 2 \) \ \( 5A - 5B = 0 \) \ Solving these, we find: \ \( A = 1 \) and \( B = 1 \) \ Thus, \( \frac{2x}{x^{2} - 25} = \frac{1}{x-5} + \frac{1}{x+5} \)

04

- Integrate using partial fractions

Now integrate each term separately: \ \( \int \frac{2 x}{x^{2}-25} d x = \int \frac{1}{x-5} d x + \int \frac{1}{x+5} d x \) \ \( = \ln|x-5| + \ln|x+5| + C \)

05

- Substitution method

Next, use the substitution \( w = x^2 - 25 \). Therefore, \( dw = 2x dx \), which gives \( dx = \frac{dw}{2x} \). \ Substitute into the integral: \ \( \int \frac{2x}{x^2 - 25} dx = \int \frac{2x}{w} \cdot \frac{dw}{2x} \) \ After simplification: \ \( = \int \frac{dw}{w} = \ln|w| + C \)

06

- Back-substitute and combine results

Since \( w = x^2 - 25 \), back-substitute \( w \): \ \( \ln|x^2 - 25| + C \) \ Note that: \ \( \ln|x - 5| + \ln|x + 5| = \ln|x^2 - 25| \) \ Thus, the results match and the final answer is: \ \( \ln|x - 5| + \ln|x + 5| + C \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fractions

In calculus, partial fractions are used to break down complex fractions into simpler parts. This technique helps to make integration easier. Consider the integral \( \int \frac{2 x}{x^{2}-25} \, dx \). Notice that the denominator can be factored as \((x-5)(x+5)\). This allows us to decompose it into simpler fractions.Using partial fractions, we can rewrite the complex fraction as \( \frac{2 x}{(x-5)(x+5)} = \frac{A}{x-5} + \frac{B}{x+5} \). Our goal is to find the values of the constants \( A \) and \( B \).By setting up the equation \( 2x = A(x+5) + B(x-5) \) and solving for \( A \) and \( B \), we find that both constants equal 1. Therefore, \( \frac{2 x}{x^{2}-25} = \frac{1}{x-5} + \frac{1}{x+5} \). Now we can integrate the simplified form easily.

Substitution Method

The substitution method is a powerful tool in calculus for simplifying integrals. When faced with a complex integral, substitution can often turn it into something more manageable.For the integral \( \int \frac{2x}{x^{2}-25} \, dx \), we use the substitution \( w = x^{2} - 25 \). This turns the integral into a form that is easier to integrate. By differentiating, we get \( dw = 2x \, dx \). Consequently, \( dx \) can be replaced by \( \frac{dw}{2x} \).Substituting back into the integral: \( \int \frac{2x}{x^2 - 25} \, dx = \int \frac{2x}{w} \, \frac{dw}{2x} \). The \( 2x \) terms cancel out, leaving us with \( \int \frac{dw}{w} \), which is ready for simple integration.

Antiderivative

An antiderivative is a function whose derivative is the given function. When we integrate a function, we are essentially finding its antiderivative.For the integral \( \int \frac{2x}{x^2 - 25} \, dx \), after using partial fractions, we broke it down to \( \int \frac{1}{x-5} \, dx + \int \frac{1}{x+5} \, dx \). The antiderivative of \( \frac{1}{x-5} \) is \( \ln|x-5| \) and the antiderivative of \( \frac{1}{x+5} \) is \( \ln|x+5| \).These results are based on the fact that the integral of \( \frac{1}{x} \) is \( \ln|x| \). Therefore, the antiderivative of our original integrand is \( \ln|x-5| + \ln|x+5| \), plus the constant of integration.

Constant of Integration

When performing indefinite integration, we always add a constant of integration, denoted by \( +C \). This constant represents all possible vertical shifts of the antiderivative, since the derivative of any constant is zero.In our case, the integral \( \int \frac{2x}{x^2 - 25} \, dx = \ln|x-5| + \ln|x+5| + C \), where \( C \) is the constant of integration.This constant is important because it acknowledges that there are infinitely many antiderivatives for any given function, each differing by a constant.

Logarithmic Integration

Logarithmic integration occurs when the integral of a function results in a logarithm. This typically happens when we're integrating functions of the form \( \frac{1}{x} \).In our example, after decomposing the integral \( \int \frac{2x}{x^2-25} \, dx \) and using the substitution method, we arrived at integrals of the form \( \int \frac{1}{x-5} \, dx \) and \( \int \frac{1}{x+5} \, dx \). Both of these are classic scenarios for logarithmic integration.The results are \( \ln|x-5| \) and \( \ln|x+5| \), showing that logarithmic functions are the antiderivatives in this context. Therefore, the final answer is \( \ln|x-5| + \ln|x+5| + C \). This demonstrates how integral calculus often leads to logarithmic functions when dealing with rational integrands.