Indefinite Integral - Some Important Formulae

• An indefinite integral is denoted by ∫f(x)dx, where f(x) is the integrand and dx is the differential of x.
• It represents the family of all antiderivatives of f(x).

Some Important Formulae

1. Power Rule:
   • ∫x^n dx = (x^(n+1))/(n+1) + C, where n ≠ -1. Example: ∫x^2 dx = (x^3)/3 + C.

2. Constant Rule:
   • ∫k dx = kx + C, where k is a constant. Example: ∫3 dx = 3x + C.

3. Sum and Difference Rule:
   • ∫(f(x) + g(x)) dx = ∫f(x)dx + ∫g(x)dx.
   • ∫(f(x) - g(x)) dx = ∫f(x)dx - ∫g(x)dx. Example: ∫(2x + 3x^2) dx = ∫2x dx + ∫3x^2 dx = x^2 + x^3 + C.

4. Reciprocal Rule:
   • ∫(1/x) dx = ln|x| + C, where x ≠ 0. Example: ∫(1/x) dx = ln|x| + C.

5. Exponential Rule:
   • ∫e^x dx = e^x + C. Example: ∫e^x dx = e^x + C.

11. Integration by Substitution:

• Integration by substitution is a technique used to simplify complicated integrals by substituting for a new variable.
• It involves making a substitution that transforms the integral into a simpler form.
• The substitution is usually chosen based on the derivative of the function inside the integral.

12. Steps for Integration by Substitution:
13. Identify a function inside the integral that can be replaced with a new variable.
14. Choose a suitable substitution such that the derivative of the new variable simplifies the integrand.
15. Rewrite the integrand in terms of the new variable.
16. Replace dx with the appropriate differential of the new variable.
17. Integrate the new function with respect to the new variable.
18. Substitute back the original variable.

13. Example 1:

• Let’s find the integral of ∫(x^2 + 1) * x dx.
• We can let u = x^2 + 1.
• Calculating du/dx, we get du = 2x dx.
• Rewriting the integrand in terms of u, we have ∫u * (du/2).
• Integrating, we get (1/2) * ∫u du = (1/2) * (u^2/2) + C.
• Substituting back u = x^2 + 1, we get the final answer as (1/2) * ((x^2 + 1)^2/2) + C.

14. Example 2:

• Let’s find the integral of ∫3x * √(x^2 + 1) dx.
• We can let u = x^2 + 1.
• Calculating du/dx, we get du = 2x dx.
• Rewriting the integrand in terms of u, we have (3/2) * ∫√(u) du.
• Integrating, we get (3/2) * (2/3) * u^(3/2) + C = u^(3/2) + C.
• Substituting back u = x^2 + 1, we get the final answer as (x^2 + 1)^(3/2) + C.

15. Integration by Parts

• Integration by parts is a technique used to evaluate the integral of the product of two functions.
• It involves using the product rule for differentiation in a reverse manner.
• The formula for integration by parts is ∫u dv = uv - ∫v du.

16. Steps for Integration by Parts:
17. Select one function to be differentiated (u) and another function to be integrated (dv).
18. Calculate the differential of u, du, and the integral of dv, v.
19. Apply the formula for integration by parts, ∫u dv = uv - ∫v du.
20. Simplify the integral on the right-hand side and evaluate it.
21. Repeat the process if necessary until the integral can be evaluated.

17. Example 1:

• Let’s find the integral of ∫x * e^x dx.
• Choosing u = x and dv = e^x dx, we have du = dx and v = ∫e^x dx = e^x.
• Applying the formula for integration by parts, we get ∫x * e^x dx = x * e^x - ∫e^x dx.
• Simplifying, we have ∫x * e^x dx = x * e^x - e^x + C.

18. Example 2:

• Let’s find the integral of ∫ln(x) dx.
• Choosing u = ln(x) and dv = dx, we have du = (1/x) dx and v = x.
• Applying the formula for integration by parts, we get ∫ln(x) dx = x * ln(x) - ∫(1/x) dx.
• Simplifying, we have ∫ln(x) dx = x * ln(x) - ∫(1/x) dx = x * ln(x) - ln(x) + C.

19. Integration of Trigonometric Functions

• Integration of trigonometric functions involves finding the antiderivatives of functions involving trigonometric functions.

20. Some Important Formulas for Integration of Trigonometric Functions:

• ∫sin(x) dx = -cos(x) + C
• ∫cos(x) dx = sin(x) + C
• ∫sec^2(x) dx = tan(x) + C
• ∫cosec^2(x) dx = -cot(x) + C
• ∫sec(x) * tan(x) dx = sec(x) + C
• ∫cosec(x) * cot(x) dx = -cosec(x) + C

21. Integration by Partial Fractions

• Integration by partial fractions is a technique used to decompose a complex rational function into simpler fractions in order to integrate it.
• It is particularly useful when integrating functions that cannot be easily integrated using other techniques.
• The decomposition involves expressing the rational function as a sum of simpler fractions with denominators that are linear or quadratic.

22. Steps for Integration by Partial Fractions

1. Factorize the denominator of the rational function.

2. Write the rational function as a sum of simpler fractions, with each fraction having a linear or quadratic denominator.

3. Assign variables (constants) to the numerators of the fractions.

4. Express the original function as an equation and equate the numerators on both sides.

5. Solve for the variables (constants) by comparing the coefficients of similar powers of the variable(s).

6. Rewrite the original function as the sum of the decomposed fractions.

7. Integrate each decomposed fraction separately.

8. Sum up the integrals of the decomposed fractions to obtain the final result.

23. Example 1:

• Let’s find the integral of ∫(x^2 - 2x - 3)/((x-1)(x+2)) dx.
• Factorizing the denominator, we have (x-1)(x+2).
• We can write the fraction using partial fractions as A/(x-1) + B/(x+2).
• Equating the numerators, we have x^2 - 2x - 3 = A(x+2) + B(x-1).
• Simplifying, we get x^2 - 2x - 3 = (A+B)x + (2A-B).
• Comparing coefficients, we have A+B = 1 and 2A-B = -3.
• Solving the equations, we get A = 1 and B = -2.
• Rewriting the original function using partial fractions, we have ∫(1/(x-1)) - 2/(x+2) dx.
• Integrating each fraction, we get ln|x-1| - 2ln|x+2| + C.
• Therefore, the final answer is ln|x-1| - 2ln|x+2| + C.

24. Example 2:

• Let’s find the integral of ∫(3x+7)/((x+1)(x+2)^2) dx.
• Factorizing the denominator, we have (x+1)(x+2)^2.
• We can write the fraction using partial fractions as A/(x+1) + B/(x+2) + C/(x+2)^2.
• Equating the numerators, we have 3x+7 = A(x+2)^2 + B(x+1)(x+2) + C(x+1).
• Simplifying, we get 3x+7 = (A+B)x^2 + (4A+3B+C)x + (4A+2B+C).
• Comparing coefficients, we have A+B = 0, 4A+3B+C = 3, and 4A+2B+C = 7.
• Solving the equations, we get A = 2, B = -2, and C = 6.
• Rewriting the original function using partial fractions, we have ∫(2/(x+1)) - 2/(x+2) + 6/(x+2)^2 dx.
• Integrating each fraction, we get 2ln|x+1| - 2ln|x+2| - 6/(x+2) + C.
• Therefore, the final answer is 2ln|x+1| - 2ln|x+2| - 6/(x+2) + C.

25. Integration of Trigonometric Substitutions

• Integration of certain types of rational functions can be simplified by using trigonometric substitutions.
• Trigonometric substitutions involve replacing the variable in the integral with a trigonometric function.
• This technique is useful for integrating functions involving the square root of a quadratic expression or the square root of the sum or difference of squares.

26. Some Common Trigonometric Substitutions:

1. For expressions involving √(a^2 - x^2):

• Let x = a sinθ, where -π/2 ≤ θ ≤ π/2.

2. For expressions involving √(x^2 + a^2):

• Let x = a tanθ, where -π/2 < θ < π/2.

3. For expressions involving √(x^2 - a^2):

• Let x = a secθ or x = -a secθ, where 0 ≤ θ < π/2 or π < θ ≤ (3π/2).

27. Example 1:

• Let’s find the integral of ∫√(x^2 + 4) dx.
• Using the substitution x = 2 tanθ, we have dx = 2 sec^2θ dθ.
• Substituting the variables, we get ∫(2 secθ)(2 sec^2θ) dθ.
• Simplifying, we have ∫4 sec^3θ dθ.
• This integral can be solved using integration by parts or other techniques specific to trigonometric functions.

28. Example 2:

• Let’s find the integral of ∫x^2√(4 - x^2) dx.
• Using the substitution x = 2 sinθ, we have dx = 2 cosθ dθ.
• Substituting the variables, we get ∫(2 sinθ)^2√(4 - (2 sinθ)^2) (2 cosθ) dθ.
• Simplifying, we have ∫4 sin^2θ cos^2θ dθ.
• This integral can be solved using trigonometric identities and formulas.

29. Improper Integrals

• An improper integral is an integral with one or both limits of integration being infinite or where the integrand itself is unbounded or goes to infinity.
• Improper integrals require special techniques to obtain a finite value.
• They can be classified into two types: Type 1 and Type 2.

30. Type 1 Improper Integrals

• Type 1 improper integrals have infinite bounds or involve integrands that are unbounded or go to infinity at one or both limits of integration.
• The integral may diverge (not have a finite value) or converge (have a finite value) depending on the behavior of the integrand.
• For Type 1 improper integrals, the following cases can arise:

1. Infinite Interval:
   • ∫f(x) dx = lim (a->-∞) ∫(a to b) f(x) dx, if the limit exists.
   • ∫f(x) dx = lim (b->+∞) ∫(a to b) f(x) dx, if the limit exists.

2. Vertical Asymptote:
   • ∫f(x) dx = lim (c->a+) ∫(c to b) f(x) dx, if the limit exists.
   • ∫f(x) dx = lim (c->a-) ∫(c to b) f(x) dx, if the limit exists.

3. Both Infinite Interval and Vertical Asymptote:
   • ∫f(x) dx = lim (a->-∞) ∫(a to c-) f(x) dx + lim (c->a+) ∫(c to b) f(x) dx, if the limits exist.

• The integrals can be evaluated using appropriate techniques and the limit is taken as necessary.