Example 1

• Consider the integral: ∫(5x^2 - 2)/(x^3 - 5x) dx

• We start by factoring the denominator: x(x^2 - 5)

• Next, we break down the rational function into partial fractions: (5x^2 - 2)/(x(x^2 - 5)) = A/x + (Bx + C)/(x^2 - 5)

• To find the values of A, B, and C, we equate the numerators:

  5x^2 - 2 = A(x^2 - 5) + (Bx + C)x

Example 1 (continued)

• We expand and simplify the equation: 5x^2 - 2 = (A + B)x^2 + Cx - 5A
• Equating the coefficients of each term, we get the following equations: A + B = 5 C = 0 -5A = -2
• Solving these equations, we find the values of A, B, and C: A = 2/5 B = 23/5 C = 0

Example 1 (continued)

• Now we can rewrite the original integral as: ∫(5x^2 - 2)/(x(x^2 - 5)) dx = ∫(2/5)(1/x) + (23/5)(x/(x^2 - 5)) dx
• We can now integrate each term separately: ∫(2/5)(1/x) dx + ∫(23/5)(x/(x^2 - 5)) dx
• The integral of (1/x) is ln|x| and the integral of (x/(x^2 - 5)) is (1/2)*ln|(x^2 - 5)|

Example 1 (continued)

• Substituting the values back, we have: (2/5)ln|x| + (23/5)(1/2)*ln|(x^2 - 5)| + C
• This is the final result of the integral.

Example 2

• Let’s consider another example: ∫(3x^3 + 2x)/(x^4 - 2x^2) dx
• Again, we start by factoring the denominator: x^2(x^2 - 2)
• Next, we break down the rational function into partial fractions: (3x^3 + 2x)/(x^2(x^2 - 2)) = A/x + B/x^2 + C/(x^2 - 2)

Example 2 (continued)

• To find the values of A, B, and C, we equate the numerators: 3x^3 + 2x = A(x^2 - 2) + Bx(x^2 - 2) + Cx^2
• Expanding and simplifying the equation: 3x^3 + 2x = (A + B)x^3 + (-2A + C)x^2 - 2A

Example 2 (continued)

• Equating the coefficients of each term, we get the following equations: A + B = 3 -2A + C = 0 -2A = 2
• Solving these equations, we find the values of A, B, and C: A = -1 B = 4 C = -2

Example 2 (continued)

• Now we can rewrite the original integral as: ∫(3x^3 + 2x)/(x^2(x^2 - 2)) dx = ∫(-1/x) + (4/x^2) + (-2/(x^2 - 2)) dx
• We can now integrate each term separately: ∫(-1/x) dx + ∫(4/x^2) dx + ∫(-2/(x^2 - 2)) dx
• The integral of (-1/x) is -ln|x|, the integral of (4/x^2) is -4/x, and the integral of (-2/(x^2 - 2)) is -2*ln|(x^2 - 2)|

Example 2 (continued)

• Substituting the values back, we have: -ln|x| - 4/x - 2*ln|(x^2 - 2)| + C
• This is the final result of the integral.

11. Partial Fractions with Repeated Factors

• Sometimes, the denominator of a rational function may have repeated factors.
• In such cases, we use partial fractions with repeated factors to simplify the expression.
• Let’s consider an example to understand this concept. Example: ∫(x^3 + 5x^2 + 4x + 1)/(x^2 + 3x + 2)^2 dx
• Here, the denominator has a repeated factor of (x^2 + 3x + 2).
• We break down the rational function into partial fractions as follows: (x^3 + 5x^2 + 4x + 1)/(x^2 + 3x + 2)^2 = A/(x^2 + 3x + 2) + B/(x^2 + 3x + 2)^2
• To find the values of A and B, we equate the numerators and solve for the coefficients.

12. Partial Fractions with Repeated Factors (continued)

• Continuing with the example: (x^3 + 5x^2 + 4x + 1)/(x^2 + 3x + 2) = A + B(x + 2)
• Expanding the numerator: x^3 + 5x^2 + 4x + 1 = A(x^2 + 3x + 2) + B(x + 2)(x^2 + 3x + 2)
• By equating the coefficients of each power of x, we can solve for A and B.
• Once we find the values of A and B, we can integrate each term separately.
• The integrals of A and B(x + 2) can be easily calculated.
• This method is useful for simplifying integrals involving rational functions with repeated factors in the denominator.

13. Limits in Integration

• In some cases, evaluating an integral using partial fractions may lead to an indeterminate form or division by zero.
• To overcome this, we use limits to find the value of the integral.
• Let’s consider an example to understand this concept. Example: ∫(x + 2)/(x - 4) dx
• Here, the denominator (x - 4) becomes zero at x = 4.
• To evaluate the integral, we split it into two parts: before the singularity and after the singularity.
• We evaluate these two parts separately using limits and then add the results.
• This method helps us handle integrals involving rational functions with singularities.

14. Limits in Integration (continued)

• Continuing with the example: ∫(x + 2)/(x - 4) dx = ∫(x + 2)/(x - 4) dx
• Since the limit as x approaches 4 from the left side is -∞, we evaluate the integral from -∞ to 4: lim(x→4-) ∫(x + 2)/(x - 4) dx
• We find the antiderivative of (x + 2)/(x - 4) and substitute the limits of integration.
• Similarly, since the limit as x approaches 4 from the right side is +∞, we evaluate the integral from 4 to +∞: lim(x→4+) ∫(x + 2)/(x - 4) dx
• We find the antiderivative of (x + 2)/(x - 4) and substitute the limits of integration.

15. Limits in Integration (continued)

• Continuing with the example: lim(x→4-) ∫(x + 2)/(x - 4) dx = lim(x→4-) [ln|x - 4| + 3ln|x + 2|] + C
• To evaluate this limit, we substitute x = 4 in the expression: lim(x→4-) [ln|x - 4| + 3ln|x + 2|] + C = -∞
• Similarly, for the other limit, we have: lim(x→4+) ∫(x + 2)/(x - 4) dx = lim(x→4+) [ln|x - 4| + 3ln|x + 2|] + C
• To evaluate this limit, we substitute x = 4 in the expression: lim(x→4+) [ln|x - 4| + 3ln|x + 2|] + C = +∞
• Adding these two limits gives us the overall result of the integral.

16. Limits in Integration (continued)

• Continuing with the example: lim(x→4-) ∫(x + 2)/(x - 4) dx = -∞ lim(x→4+) ∫(x + 2)/(x - 4) dx = +∞
• Adding these two limits, we get the final result of the integral: ∫(x + 2)/(x - 4) dx = -∞ + ∞ = undefined
• In this case, the integral does not converge to a finite value due to the singularity at x = 4.
• Hence, the integral is considered undefined.
• Limits are a powerful tool to evaluate integrals involving such cases.

17. Integration with Logarithmic Functions

• Logarithmic functions often appear in integrals involving partial fractions.
• To integrate a logarithmic function, we use the properties of logarithms.
• Let’s look at an example to understand this concept. Example: ∫(3x + 1)/x ln(x) dx
• Here, the integral involves a logarithmic function.
• We split the integral into two parts: (3x/x) and (1/x) ln(x).
• We can integrate (3x/x) as 3 and use the property ln(ab) = ln(a) + ln(b) to integrate (1/x) ln(x).
• This approach helps us simplify integrals involving logarithmic functions.

18. Integration with Logarithmic Functions (continued)

• Continuing with the example: ∫(3x + 1)/x ln(x) dx = ∫3 dx + ∫(1/x) ln(x) dx
• Integrating 3 with respect to x gives us 3x.
• For the second term, we use the property ln(ab) = ln(a) + ln(b) to rewrite it as ∫(ln(x) + ln(x)) (1/x) dx.
• Simplifying further, we have ∫ln(x)(1/x) dx + ∫ln(x)(1/x) dx.
• We can now integrate each term separately.
• This method helps us handle integrals involving logarithmic functions.

19. Integration with Logarithmic Functions (continued)

• Continuing with the example: ∫(3x + 1)/x ln(x) dx = 3x + ∫ln(x)(1/x) dx + ∫ln(x)(1/x) dx
• The integral of (ln(x)(1/x)) can be evaluated using the property ∫ln(x) dx = x(ln|x| - 1) + C.
• Hence, we have: ∫ln(x)(1/x) dx = x(ln|x| - 1) + C
• Substituting this result back into the equation, we get: ∫(3x + 1)/x ln(x) dx = 3x + x(ln|x| - 1) + C
• This is the final result of the integral.

20. Summary

• In this lecture, we covered the concept of partial fractions and limits in integration.
• Partial fractions are used to simplify integrals involving rational functions.
• If the denominator has repeated factors, we use partial fractions with repeated factors.
• When evaluating integrals leads to indeterminate forms or division by zero, we use limits to find the value of the integral.
• Logarithmic functions often appear in integrals involving partial fractions, and we can integrate them using the properties of logarithms.
• Understanding these concepts is crucial for solving integration problems effectively.

Slide 21

• Integration using partial fractions can be used to solve a wide range of integrals with rational functions.
• The process involves breaking down the rational function into simpler fractions.
• By equating the numerators and solving for the coefficients, we can determine the values of the partial fractions.
• The integral can then be rewritten as the sum of the integrals of each partial fraction.
• This method allows us to handle complex integrals more easily.

Slide 22

• Let’s consider an example: ∫(x^2 + 2x + 1)/(x^3 + 2x^2 + x) dx
• First, we factorize the denominator: x(x^2 + 2x + 1)
• Next, we perform partial fraction decomposition: (x^2 + 2x + 1)/(x(x^2 + 2x + 1)) = A/x + B/(x^2 + 2x + 1)
• To find the values of A and B, we equate the numerators: x^2 + 2x + 1 = A(x^2 + 2x + 1) + Bx

Slide 23

• Expanding and simplifying the equation: x^2 + 2x + 1 = A(x^2 + 2x + 1) + Bx x^2 + 2x + 1 = Ax^2 + 2Ax + A + Bx
• Equating the coefficients of each term, we get the following equations: A = 1 2A + B = 2
• Solving these equations, we find the values of A and B: A = 1 B = 0

Slide 24

• Now we can rewrite the original integral as: ∫(x^2 + 2x + 1)/(x(x^2 + 2x + 1)) dx = ∫(1/x) + (0)/(x^2 + 2x + 1) dx
• We can now integrate each term separately: ∫(1/x) dx + ∫(0)/(x^2 + 2x + 1) dx
• The integral of (1/x) is ln|x| and the integral of (0) is 0.

Slide 25

• Substituting the values back, we have: ∫(1/x) dx + ∫(0)/(x^2 + 2x + 1) dx = ln|x| + C
• This is the final result of the integral.
• Integration using partial fractions simplifies the process of solving rational function integrals.
• It allows us to apply known integration techniques to individual terms.

Slide 26

• Sometimes, integrals involving rational functions can lead to indeterminate forms or division by zero.
• In such cases, limits can be used to evaluate the integral.
• Let’s consider an example: ∫(1/x) dx from -1 to 1
• The denominator of the rational function becomes zero at x = 0.
• To evaluate the integral, we split it into two parts: before and after the singularity.

Slide 27

• The integral from -1 to 1 can be split into two parts: ∫(1/x) dx = ∫(1/x) dx from -1 to 0 + ∫(1/x) dx from 0 to 1
• Evaluating the integral from -1 to 0: ∫(1/x) dx from -1 to 0 = lim(x→0-) ∫(1/x) dx
• This limit evaluates to -∞ because the function diverges as x approaches 0 from the left side.

Slide 28

• Evaluating the integral from 0 to 1: ∫(1/x) dx from 0 to 1 = lim(x→0+) ∫(1/x) dx
• This limit evaluates to +∞ because the function diverges as x approaches 0 from the right side.
• Adding the two limits, we get the overall result of the integral.

Slide 29

• The result of the integral: ∫(1/x) dx from -1 to 1 is undefined.
• The singularity at x = 0 causes the integral to diverge.
• Limits provide a powerful tool for evaluating integrals involving rational functions with singularities.
• It helps us handle undefined or infinite integrals effectively.

Slide 30

• In summary, integration using partial fractions is a valuable technique for solving integrals with rational functions.
• It involves breaking down the function into simpler fractions, solving for the coefficients, and integrating each term individually.
• Limits are used to evaluate integrals that lead to indeterminate forms or division by zero.
• They help us handle singularities and undefined integrals.
• Understanding these concepts is essential for mastering integral calculus.