3.5 Q-16

Question Statement

Evaluate the integral:

$$\int \frac{x^{2}-6x^{2}+25}{(x+1)^{2}(x-2)^{2}} , dx$$

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Background and Explanation

This is an example of a rational function that can be solved using partial fraction decomposition. The integrand involves a fraction where the denominator is a product of squared binomials, which suggests that we can decompose it into simpler fractions for easier integration.

We need to express the given function as a sum of partial fractions, each of which will be easier to integrate individually.

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Solution

Step 1: Set up the partial fraction decomposition

We aim to decompose the integrand into the form:

$$\frac{x^2 - 6x^2 + 25}{(x+1)^2(x-2)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2} + \frac{D}{(x-2)^2}$$

Here, A, B, C, and D are constants that we need to determine.

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Step 2: Multiply both sides by the common denominator

Multiply both sides of the equation by $(x+1)^2(x-2)^2$ to eliminate the denominators:

$$x^2 - 6x^2 + 25 = A(x+1)(x-2)^2 + B(x-2)^2 + C(x-2)(x+1)^2 + D(x+1)^2$$

This results in the equation:

$$x^2 - 6x^2 + 25 = A(x+1)(x-2)^2 + B(x-2)^2 + C(x-2)(x+1)^2 + D(x+1)^2 \tag{ii}$$

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Step 3: Substitute convenient values of $x$ to solve for constants

For $x = -1$:

Substituting $x = -1$ into equation (ii), we get:

$$(-1)^2 - 6(-1)^2 + 25 = B(-1-2)^2$$

Simplifying:

$$1 - 6 + 25 = B(3)^2 \Rightarrow 9B = 18 \Rightarrow B = 2$$

For $x = 2$:

Substituting $x = 2$ into equation (ii), we get:

$$(2)^2 - 6(2)^2 + 25 = D(2+1)^2$$

Simplifying:

$$4 - 24 + 25 = D(3)^2 \Rightarrow 9D = 9 \Rightarrow D = 1$$

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Step 4: Compare coefficients of powers of $x$

We now compare the coefficients of $x^3$ and constant terms to solve for $A$ and $C$.

From the comparison of $x^3$-terms:

$$1 = A + C \Rightarrow A = 1 - C$$

Next, comparing the constant term:

$$25 = 4A + 4B - 2C + D$$

Substituting $B = 2$ and $D = 1$:

$$25 = 4A + 8 - 2C + 1 \Rightarrow 25 = 4A - 2C + 9$$

Simplifying:

$$25 - 9 = 4A - 2C \Rightarrow 16 = 4A - 2C$$

Now substitute $A = 1 - C$:

$$16 = 4(1 - C) - 2C \Rightarrow 16 = 4 - 4C - 2C \Rightarrow 16 = 4 - 6C$$

Solving for $C$:

$$12 = -6C \Rightarrow C = -2$$

Now substitute $C = -2$ into $A = 1 - C$:

$$A = 1 - (-2) = 3$$

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Step 5: Write the partial fractions

Now we can write the integral as:

$$\int \frac{x^2 - 6x^2 + 25}{(x+1)^2(x-2)^2} , dx = \int \left[ \frac{3}{x+1} + \frac{2}{(x+1)^2} - \frac{2}{x-2} + \frac{1}{(x-2)^2} \right] , dx$$

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Step 6: Integrate each term

Now we can integrate each term individually:

$$\int \frac{3}{x+1} , dx = 3 \ln |x+1|$$ $$\int \frac{2}{(x+1)^2} , dx = -\frac{2}{x+1}$$ $$\int \frac{-2}{x-2} , dx = -2 \ln |x-2|$$ $$\int \frac{1}{(x-2)^2} , dx = \frac{1}{x-2}$$

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Step 7: Combine the results

Combining the results of the individual integrals, we get:

$$3 \ln |x+1| - \frac{2}{x+1} - 2 \ln |x-2| + \frac{1}{x-2} + C$$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: This method allows us to decompose complex rational expressions into simpler fractions that are easier to integrate.
• Basic Integration: The integral of $\frac{1}{x}$ is $\ln|x|$, and the integral of $\frac{1}{(x-a)^2}$ is $-\frac{1}{x-a}$.

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Summary of Steps

1. Set up the partial fraction decomposition.
2. Multiply both sides by $(x+1)^2(x-2)^2$ to eliminate denominators.
3. Substitute values of $x$ (like $x = -1$ and $x = 2$) to solve for constants $A$, $B$, $C$, and $D$.
4. Compare coefficients of powers of $x$ to find any remaining constants.
5. Substitute the constants back into the partial fractions.
6. Integrate each term.
7. Combine the results and include the constant of integration.