Evaluate The Integral. Express Numbers In Exact Form. Use Symbolic Notation And Fractions Where Needed. Use \[$ C \$\] For The Arbitrary Constant. Absorb Into \[$ C \$\] As Much As Possible.$\[ \int \frac{x^2+x}{(x-9)(x-10)^2} \, Dx

Introduction

In this article, we will evaluate the integral of a rational function using symbolic notation and fractions. The integral is given as $\int \frac{x^2+x}{(x-9)(x-10)^2} \, dx$. We will use the method of partial fractions to decompose the rational function into simpler fractions, which can then be integrated separately.

Partial Fraction Decomposition

To decompose the rational function into partial fractions, we first need to factor the denominator. The denominator is already factored as $(x-9)(x-10)^2$. We can now write the rational function as:

$$\frac{x^2+x}{(x-9)(x-10)^2} = \frac{A}{x-9} + \frac{B}{x-10} + \frac{C}{(x-10)^2}$$

where $A$, $B$, and $C$ are constants to be determined.

Finding the Constants

To find the constants $A$, $B$, and $C$, we multiply both sides of the equation by the common denominator $(x-9)(x-10)^2$. This gives us:

$$x^2+x = A(x-10)^2 + B(x-9)(x-10) + C(x-9)$$

We can now expand the right-hand side of the equation and equate the coefficients of like terms.

Equating Coefficients

Expanding the right-hand side of the equation, we get:

$$x^2+x = A(x^2-20x+100) + B(x^2-19x+90) + C(x-9)$$

Equating the coefficients of like terms, we get:

$$A+B = 1$$

$$-20A-19B+C = 1$$

$$100A+90B-9C = 0$$

Solving the System of Equations

We can now solve the system of equations to find the values of $A$, $B$, and $C$. We can start by solving the first equation for $B$:

$$B = 1-A$$

Substituting this expression for $B$ into the second equation, we get:

$$-20A-19(1-A)+C = 1$$

Simplifying this equation, we get:

$$-20A-19+19A+C = 1$$

Combine like terms:

$$-A+C = 20$$

Continuing the Solution

We can now substitute the expression for $B$ into the third equation:

$$100A+90(1-A)-9C = 0$$

Simplifying this equation, we get:

$$100A+90-90A-9C = 0$$

Combine like terms:

$$10A-9C = -90$$

Solving for A and C

We can now solve the system of equations to find the values of $A$ and $C$. We can start by solving the first equation for $C$:

$$C = -A+20$$

Substituting this expression for $C$ into the second equation, we get:

$$-A+(-A+20) = 20$$

Simplifying this equation, we get:

$$-2A+20 = 20$$

Subtract 20 from both sides:

$$-2A = 0$$

Divide both sides by -2:

$$A = 0$$

Finding the Value of C

Now that we have found the value of $A$, we can find the value of $C$:

$$C = -A+20$$

Substituting $A=0$ into this equation, we get:

$$C = -0+20$$

Simplifying this equation, we get:

$$C = 20$$

Finding the Value of B

Now that we have found the values of $A$ and $C$, we can find the value of $B$:

$$B = 1-A$$

Substituting $A=0$ into this equation, we get:

$$B = 1-0$$

Simplifying this equation, we get:

$$B = 1$$

Writing the Partial Fraction Decomposition

We can now write the partial fraction decomposition of the rational function:

$$\frac{x^2+x}{(x-9)(x-10)^2} = \frac{0}{x-9} + \frac{1}{x-10} + \frac{20}{(x-10)^2}$$

Integrating the Partial Fractions

We can now integrate the partial fractions separately:

$$\int \frac{x^2+x}{(x-9)(x-10)^2} \, dx = \int \frac{0}{x-9} \, dx + \int \frac{1}{x-10} \, dx + \int \frac{20}{(x-10)^2} \, dx$$

Evaluating the Integrals

We can now evaluate the integrals separately:

$$\int \frac{0}{x-9} \, dx = 0$$

$$\int \frac{1}{x-10} \, dx = \ln|x-10|+C_1$$

$$\int \frac{20}{(x-10)^2} \, dx = -\frac{20}{x-10}+C_2$$

Combining the Results

We can now combine the results of the integrals:

$$\int \frac{x^2+x}{(x-9)(x-10)^2} \, dx = 0 + \ln|x-10|+C_1 -\frac{20}{x-10}+C_2$$

Simplifying the Result

We can now simplify the result by combining the constants:

$$\int \frac{x^2+x}{(x-9)(x-10)^2} \, dx = \ln|x-10| -\frac{20}{x-10}+C$$

where $C=C_1+C_2$ is an arbitrary constant.

Conclusion

In this article, we evaluated the integral of a rational function using symbolic notation and fractions. We used the method of partial fractions to decompose the rational function into simpler fractions, which can then be integrated separately. We found the values of the constants $A$, $B$, and $C$ by equating the coefficients of like terms and solving the resulting system of equations. We then integrated the partial fractions separately and combined the results to obtain the final answer.

Introduction

In our previous article, we evaluated the integral of a rational function using symbolic notation and fractions. We used the method of partial fractions to decompose the rational function into simpler fractions, which can then be integrated separately. In this article, we will answer some common questions that readers may have about the process of evaluating the integral of a rational function.

Q: What is the method of partial fractions?

A: The method of partial fractions is a technique used to decompose a rational function into simpler fractions. It involves expressing the rational function as a sum of simpler fractions, each with a polynomial numerator and a polynomial denominator.

Q: How do I know when to use the method of partial fractions?

A: You should use the method of partial fractions when you are trying to integrate a rational function and the denominator is a product of linear factors. This is because the method of partial fractions allows you to express the rational function as a sum of simpler fractions, each of which can be integrated separately.

Q: What are the steps involved in using the method of partial fractions?

A: The steps involved in using the method of partial fractions are:

1. Factor the denominator of the rational function.
2. Express the rational function as a sum of simpler fractions, each with a polynomial numerator and a polynomial denominator.
3. Integrate each of the simpler fractions separately.
4. Combine the results of the integrals to obtain the final answer.

Q: How do I find the values of the constants in the partial fraction decomposition?

A: To find the values of the constants in the partial fraction decomposition, you need to equate the coefficients of like terms and solve the resulting system of equations. This involves multiplying both sides of the equation by the common denominator and equating the coefficients of like terms.

Q: What is the significance of the arbitrary constant in the final answer?

A: The arbitrary constant in the final answer is a constant that can take on any value. It is used to represent the fact that the integral is not defined up to a constant multiple.

Q: Can I use the method of partial fractions to integrate rational functions with non-linear denominators?

A: No, the method of partial fractions can only be used to integrate rational functions with linear denominators. If the denominator is non-linear, you will need to use a different method, such as substitution or integration by parts.

Q: How do I know when to use substitution or integration by parts instead of the method of partial fractions?

A: You should use substitution or integration by parts instead of the method of partial fractions when the denominator is non-linear or when the rational function is not easily decomposed into simpler fractions.

Q: Can I use the method of partial fractions to integrate rational functions with complex denominators?

A: No, the method of partial fractions can only be used to integrate rational functions with real denominators. If the denominator is complex, you will need to use a different method, such as contour integration or the residue theorem.

Q: How do I know when to use contour integration or the residue theorem instead of the method of partial fractions?

A: You should use contour integration or the residue theorem instead of the method of partial fractions when the denominator is complex or when the rational function is not easily decomposed into simpler fractions.

Conclusion

In this article, we answered some common questions that readers may have about the process of evaluating the integral of a rational function. We discussed the method of partial fractions, the steps involved in using it, and the significance of the arbitrary constant in the final answer. We also discussed when to use substitution or integration by parts instead of the method of partial fractions, and when to use contour integration or the residue theorem instead of the method of partial fractions.