Evaluate The Integral:${ \int \frac{(-x) , D X}{-4 X^2+8} }$

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Introduction

In this article, we will delve into the world of calculus and evaluate the integral of a rational function. Rational functions are a type of function that can be expressed as the ratio of two polynomials. They are commonly encountered in various fields of mathematics, physics, and engineering. The integral of a rational function can be evaluated using various techniques, including partial fractions, substitution, and integration by parts.

The Integral to be Evaluated

The integral we will be evaluating is:

∫(−x) dx−4x2+8\int \frac{(-x) \, d x}{-4 x^2+8}

This is a rational function with a quadratic denominator. To evaluate this integral, we will use the technique of partial fractions.

Partial Fractions

Partial fractions is a technique used to break down a rational function into simpler fractions. This technique is useful when the denominator of the rational function can be factored into linear or quadratic factors. In this case, the denominator can be factored as:

−4x2+8=−4(x2−2)-4 x^2+8 = -4(x^2-2)

We can now rewrite the integral as:

∫(−x) dx−4(x2−2)\int \frac{(-x) \, d x}{-4(x^2-2)}

Breaking Down the Rational Function

To break down the rational function, we will use the following partial fraction decomposition:

(−x)−4(x2−2)=Ax−2+Bx+2\frac{(-x)}{-4(x^2-2)} = \frac{A}{x-\sqrt{2}} + \frac{B}{x+\sqrt{2}}

where A and B are constants to be determined.

Determining the Constants

To determine the constants A and B, we will multiply both sides of the equation by the common denominator:

−x=A(x+2)+B(x−2)-x = A(x+\sqrt{2}) + B(x-\sqrt{2})

We can now expand the right-hand side of the equation:

−x=Ax+A2+Bx−B2-x = Ax + A\sqrt{2} + Bx - B\sqrt{2}

Combining like terms, we get:

−x=(A+B)x+(A−B)2-x = (A+B)x + (A-B)\sqrt{2}

Equating the coefficients of x and the constant terms, we get:

A+B=0andA−B=−1A+B = 0 \quad \text{and} \quad A-B = -1

Solving these equations simultaneously, we get:

A=−12andB=12A = -\frac{1}{2} \quad \text{and} \quad B = \frac{1}{2}

Evaluating the Integral

Now that we have determined the constants A and B, we can rewrite the integral as:

∫(−x) dx−4(x2−2)=∫(−12x−2+12x+2) dx\int \frac{(-x) \, d x}{-4(x^2-2)} = \int \left( \frac{-\frac{1}{2}}{x-\sqrt{2}} + \frac{\frac{1}{2}}{x+\sqrt{2}} \right) \, d x

We can now evaluate the integral using the technique of substitution:

∫(−12x−2+12x+2) dx=−12∫1x−2 dx+12∫1x+2 dx\int \left( \frac{-\frac{1}{2}}{x-\sqrt{2}} + \frac{\frac{1}{2}}{x+\sqrt{2}} \right) \, d x = -\frac{1}{2} \int \frac{1}{x-\sqrt{2}} \, d x + \frac{1}{2} \int \frac{1}{x+\sqrt{2}} \, d x

Evaluating the Substitution

To evaluate the substitution, we will use the following substitution:

u=x−2andv=x+2u = x-\sqrt{2} \quad \text{and} \quad v = x+\sqrt{2}

We can now rewrite the integral as:

−12∫1u du+12∫1v dv-\frac{1}{2} \int \frac{1}{u} \, d u + \frac{1}{2} \int \frac{1}{v} \, d v

We can now evaluate the integral using the technique of integration by parts:

−12∫1u du+12∫1v dv=−12ln⁡∣u∣+12ln⁡∣v∣+C-\frac{1}{2} \int \frac{1}{u} \, d u + \frac{1}{2} \int \frac{1}{v} \, d v = -\frac{1}{2} \ln |u| + \frac{1}{2} \ln |v| + C

Substituting Back

We can now substitute back to get:

−12ln⁡∣x−2∣+12ln⁡∣x+2∣+C-\frac{1}{2} \ln |x-\sqrt{2}| + \frac{1}{2} \ln |x+\sqrt{2}| + C

Simplifying the Expression

We can now simplify the expression by combining the logarithms:

−12ln⁡∣x−2∣+12ln⁡∣x+2∣+C=12ln⁡∣x+2x−2∣+C-\frac{1}{2} \ln |x-\sqrt{2}| + \frac{1}{2} \ln |x+\sqrt{2}| + C = \frac{1}{2} \ln \left| \frac{x+\sqrt{2}}{x-\sqrt{2}} \right| + C

Conclusion

In this article, we evaluated the integral of a rational function using the technique of partial fractions. We broke down the rational function into simpler fractions and determined the constants A and B. We then evaluated the integral using the technique of substitution and integration by parts. The final answer is:

12ln⁡∣x+2x−2∣+C\frac{1}{2} \ln \left| \frac{x+\sqrt{2}}{x-\sqrt{2}} \right| + C

This is the final answer to the integral.

Final Answer

The final answer is 12ln⁡∣x+2x−2∣+C\boxed{\frac{1}{2} \ln \left| \frac{x+\sqrt{2}}{x-\sqrt{2}} \right| + C}.

References

  • [1] "Calculus" by Michael Spivak
  • [2] "Calculus" by James Stewart
  • [3] "Partial Fractions" by Wolfram MathWorld

Additional Resources

  • [1] "Calculus" by Khan Academy
  • [2] "Partial Fractions" by MIT OpenCourseWare
  • [3] "Calculus" by Coursera

Introduction

In our previous article, we evaluated the integral of a rational function using the technique of partial fractions. In this article, we will answer some common questions that readers may have about the integral.

Q: What is the integral of a rational function?

A: The integral of a rational function is a mathematical operation that involves finding the antiderivative of a rational function. A rational function is a function that can be expressed as the ratio of two polynomials.

Q: How do I evaluate the integral of a rational function?

A: To evaluate the integral of a rational function, you can use the technique of partial fractions. This involves breaking down the rational function into simpler fractions and then evaluating the integral of each fraction separately.

Q: What is partial fractions?

A: Partial fractions is a technique used to break down a rational function into simpler fractions. This involves expressing the rational function as a sum of simpler fractions, each with a linear or quadratic denominator.

Q: How do I determine the constants in partial fractions?

A: To determine the constants in partial fractions, you can use the following steps:

  1. Multiply both sides of the equation by the common denominator.
  2. Expand the right-hand side of the equation.
  3. Combine like terms.
  4. Equate the coefficients of x and the constant terms.
  5. Solve the resulting equations simultaneously.

Q: What is the final answer to the integral?

A: The final answer to the integral is:

12ln⁡∣x+2x−2∣+C\frac{1}{2} \ln \left| \frac{x+\sqrt{2}}{x-\sqrt{2}} \right| + C

Q: Can I use other techniques to evaluate the integral?

A: Yes, you can use other techniques to evaluate the integral, such as substitution and integration by parts. However, partial fractions is a powerful technique that can be used to evaluate many types of integrals.

Q: What are some common applications of the integral of a rational function?

A: The integral of a rational function has many applications in mathematics, physics, and engineering. Some common applications include:

  • Finding the area under a curve
  • Finding the volume of a solid
  • Finding the work done by a force
  • Finding the energy of a system

Q: Can I use the integral of a rational function to solve real-world problems?

A: Yes, you can use the integral of a rational function to solve many real-world problems. For example, you can use it to find the area under a curve, the volume of a solid, or the work done by a force.

Q: What are some common mistakes to avoid when evaluating the integral of a rational function?

A: Some common mistakes to avoid when evaluating the integral of a rational function include:

  • Not factoring the denominator
  • Not using partial fractions
  • Not determining the constants correctly
  • Not simplifying the expression

Conclusion

In this article, we answered some common questions that readers may have about the integral of a rational function. We also provided some tips and tricks for evaluating the integral and some common applications of the integral.

Final Answer

The final answer to the integral is 12ln⁡∣x+2x−2∣+C\boxed{\frac{1}{2} \ln \left| \frac{x+\sqrt{2}}{x-\sqrt{2}} \right| + C}.

References

  • [1] "Calculus" by Michael Spivak
  • [2] "Calculus" by James Stewart
  • [3] "Partial Fractions" by Wolfram MathWorld

Additional Resources

  • [1] "Calculus" by Khan Academy
  • [2] "Partial Fractions" by MIT OpenCourseWare
  • [3] "Calculus" by Coursera