Integral Calculus - Integration using Partial fraction and Limit

  • To solve integrals involving rational functions, we use the method of partial fractions.
  • This method involves breaking down the rational function into simpler fractions.
  • The method also involves evaluating the limit, as required.
  • Let’s look at some examples to understand this concept better.

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.
  1. 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.
  1. 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.
  1. 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.
  1. 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.
  1. 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.
  1. 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.
  1. 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.
  1. 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.
  1. 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.
  1. 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.