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Quantitative Aptitude · Differential and Integral Calculus

Integration Basics and Standard Formulas for CA Foundation

Updated 1 October 2026 · Fact-checked

Integration is the reverse of differentiation. If d/dx [F(x)] = f(x), then ∫ f(x) dx = F(x) + C. To solve a question, rewrite the function as a sum of powers of x, eᵃˣ or 1/x, apply the standard formula to each term, and add the constant C.

Understand Integration Basics and Standard Formulas

Differentiation takes a function and gives its rate of change. Integration goes the other way. You are given the rate of change and you must find the original function. That is why it is called anti-differentiation.

The result of an indefinite integral is not one function but a family. d/dx (x² + 5) and d/dx (x² − 9) both equal 2x. So ∫ 2x dx = x² + C, where C is the constant of integration. It stands for any constant that vanished when differentiating.

In business, you integrate marginal cost to get total cost, or marginal revenue to get total revenue. The constant C is then found from a given condition, such as fixed cost.

Most CA Foundation questions need only a few standard results: the power rule, ∫ 1/x dx, and ∫ eˣ dx. Linear rules let you integrate term by term. Before integrating, simplify first. Expand brackets, split fractions and convert roots to powers.

For a slightly harder function, such as (2x + 3)⁴ or e⁵ˣ, the substitution method lets you reduce it to a standard form. For a linear inside part (ax + b), there is a quick shortcut: divide by a.

Key formulas to remember

Power rule
∫ xⁿ dx = xⁿ⁺¹ ÷ (n + 1) + C
Valid only for n ≠ −1.
Reciprocal rule
∫ (1/x) dx = log x + C (for x > 0; in general log |x| + C)
This is the case n = −1. Here log means natural logarithm (base e).
Exponential rule
∫ eˣ dx = eˣ + C
eˣ is unchanged by integration.
Exponential with constant base
∫ aˣ dx = aˣ ÷ log a + C (a > 0, a ≠ 1)
Log is natural log.
Constant
∫ k dx = kx + C
Special case of the power rule with n = 0.
Constant multiple rule
∫ k·f(x) dx = k ∫ f(x) dx
Take the constant outside.
Sum and difference rule
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
Integrate term by term. There is no such rule for products or quotients.
Linear inside function
∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ ÷ [a(n + 1)] + C (n ≠ −1); ∫ eᵃˣ⁺ᵇ dx = eᵃˣ⁺ᵇ ÷ a + C; ∫ 1/(ax + b) dx = (1/a) log |ax + b| + C
Follows from substitution t = ax + b. Divide by a.
Check by differentiation
d/dx [∫ f(x) dx] = f(x)
Use it to verify an answer or to eliminate options.

How to solve Integration Basics and Standard Formulas questions

Use this order for any basic integration question. It also works when the options look similar.

  1. 1Simplify the integrand. Expand brackets, split the fraction term by term and write roots as powers, for example √x = x^(1/2) and 1/x³ = x⁻³.
  2. 2Identify each term's type: power of x, 1/x, eˣ, aˣ or a constant.
  3. 3Apply the matching standard formula to each term. For xⁿ, add 1 to the power and divide by the new power.
  4. 4If the inside is linear, like (ax + b), integrate as if it were x and then divide by a. For other inside functions, use substitution t = inside part.
  5. 5Combine all terms and add one constant C. In a definite condition question, use the given condition to find C.
  6. 6Differentiate your answer mentally and check it matches the original integrand.

Quickest way: Option elimination by differentiation

When to use it: Use this in MCQs when the function is not trivial or you fear a sign or coefficient slip. Differentiating four options is often faster than integrating carefully.

  1. Write the integrand and note its highest power and any e or log term.
  2. Integrate the first term only, and drop options that have a different first term.
  3. Differentiate the surviving options. The one that returns the integrand is correct.
  4. For (ax + b)ⁿ, remember the answer has a in the denominator. Options without 1/a are wrong.
  5. If options differ only by a constant, they are equal as indefinite integrals. Such a set should not appear. Pick the one that matches the given condition if any.
  6. Do not spend more than about a minute. If the form is unfamiliar, skip it, since each wrong answer costs 0.25 marks.

Common mistakes in Integration Basics and Standard Formulas

  • Forgetting the constant C in an indefinite integral.

    Students focus only on the formula and treat the answer as one function.

    Fix: Always write + C at the end. In MCQs, the option with + C is the expected one.

  • Using the power rule on 1/x, giving x⁰ ÷ 0.

    The rule xⁿ⁺¹ ÷ (n + 1) is applied blindly when n = −1.

    Fix: Check the power first. If it is −1, the answer is log x + C.

  • Not dividing by a in ∫ (ax + b)ⁿ dx or ∫ e^(ax) dx.

    Students treat the inside like a plain x.

    Fix: After integrating, divide by the coefficient of x. Check by differentiating: the chain rule brings a back.

  • Integrating a product or quotient factor by factor.

    Students overextend the sum rule.

    Fix: There is no product rule like that. Expand or split first, for example (x + 1)² = x² + 2x + 1, then integrate each term.

  • Writing ∫ aˣ dx = aˣ, or ∫ eˣ dx = x eˣ.

    Confusing the rule for eˣ with the rule for general base or with the power rule.

    Fix: Only eˣ stays unchanged. For aˣ, divide by log a. Never apply the power rule to a variable exponent.

  • Mistakes in fractional and negative powers, such as ∫ √x dx = x^(3/2) without dividing by 3/2.

    Adding 1 to the power is done, but the division is skipped or inverted.

    Fix: Divide by the new power, which means multiplying by its reciprocal. So ∫ x^(1/2) dx = (2/3) x^(3/2) + C.

Worked examples

Example 1

∫ (3x² + 4x − 5) dx equals: (A) x³ + 2x² − 5x + C (B) 6x + 4 + C (C) 3x³ + 4x² − 5x + C (D) x³ + 4x² − 5x + C

Show the solution
  1. Integrate term by term.
  2. ∫ 3x² dx = 3 · x³ ÷ 3 = x³.
  3. ∫ 4x dx = 4 · x² ÷ 2 = 2x².
  4. ∫ (−5) dx = −5x.
  5. Add the constant: x³ + 2x² − 5x + C.
  6. Check: differentiate to get 3x² + 4x − 5. This matches.

Answer: (A) x³ + 2x² − 5x + C

Example 2

∫ (x + 1/x + eˣ) dx equals: (A) x² + log x + eˣ + C (B) x²/2 + log x + eˣ + C (C) x²/2 − 1/x² + eˣ + C (D) x²/2 + log x + x eˣ + C

Show the solution
  1. Split into three terms.
  2. ∫ x dx = x²/2.
  3. ∫ (1/x) dx = log x.
  4. ∫ eˣ dx = eˣ.
  5. Sum: x²/2 + log x + eˣ + C.
  6. Option A fails because ∫ x dx needs division by 2. Option C uses the power rule wrongly on 1/x. Option D has x eˣ, which is wrong.

Answer: (B) x²/2 + log x + eˣ + C

Example 3

The marginal cost of a firm is MC = 6x + 4 and the fixed cost is ₹10. The total cost C(x) is: (A) 3x² + 4x + 10 (B) 6x² + 4x + 10 (C) 3x² + 4x (D) 3x² + 4x + 6

Show the solution
  1. Total cost is the integral of marginal cost: C(x) = ∫ (6x + 4) dx.
  2. ∫ 6x dx = 3x² and ∫ 4 dx = 4x, so C(x) = 3x² + 4x + K.
  3. Fixed cost is the cost when x = 0. So C(0) = K = 10.
  4. Therefore C(x) = 3x² + 4x + 10.
  5. Check: differentiating gives 6x + 4, which is the marginal cost.

Answer: (A) 3x² + 4x + 10

Exam tips

  • Questions are mostly direct. Scan the options first, since they usually differ in one coefficient or one missing term, and verify by differentiating.
  • Watch for linear inside functions like (2x + 3)⁴ or e^(5x). The wrong options usually have the missing 1/a factor.
  • Rewrite roots and denominators as powers before integrating. Many errors come from skipped simplification, not from the formula.
  • In cost and revenue questions, use the given condition (fixed cost, or revenue at zero output) to find C. Do not leave it as C.
  • Each wrong answer costs 0.25 marks, so skip a question if you cannot check the answer in about a minute.

Practice questions from Differential and Integral Calculus

Integration Basics and Standard Formulas: frequently asked questions

What is the difference between indefinite and definite integration?

An indefinite integral gives a family of functions and includes the constant C. A definite integral has limits, gives a number, and does not need C. Definite integrals are a separate topic.

Do I need to memorise many integration formulas for CA Foundation?

No. The core set is small: the power rule, 1/x, eˣ, aˣ, the constant rule and the linear rules. Knowing how to apply them to the linear form (ax + b) covers most questions.

How does integration by substitution work?

You replace a part of the function with a new variable t, and replace dx using dt. Then you integrate in t and put the original expression back. For a linear inside part ax + b, set t = ax + b, so dx = dt ÷ a.

Why is the constant C added?

Constants disappear when you differentiate. So when you reverse the process, you cannot know which constant was there. C represents that unknown constant.