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MATH1011 Chap.11 Antiderivatives and Integration Methods

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Chapter 11 of 12 · MATH1011

Antiderivatives and Integration Methods

Integration starts by reversing differentiation

An antiderivative of f is a function whose derivative is f. Since every constant disappears under differentiation, an indefinite integral represents a whole family and must include an arbitrary constant.

Different-looking answers can describe the same family when they differ only by a constant, so checking by differentiation is often faster than forcing two forms to look identical.

Linearity lets you integrate a sum term by term and pull constant multipliers outside. The power rule raises the exponent by one and divides by the new exponent, except at exponent minus 1, where the logarithm appears instead.

Before applying any rule, rewrite roots and reciprocals as powers so the exponent you are changing is visible.

The integrand usually advertises the method

Substitution reverses the chain rule. Look for an inner expression and a nearby factor proportional to its derivative.

Name the new variable, replace the differential as well as the expression, integrate in the new variable, and then return to the original variable. A substitution that leaves stray x terms has not completed the transformation.

Integration by parts reverses the product rule. Choose one factor to differentiate because it becomes simpler and the other to integrate because it remains manageable.

The formula trades the original product for a boundary term minus a new integral. If the new integral is harder, reconsider the choice before expanding more algebra.

Algebra comes before calculus for rational functions

Long division is required when the numerator's degree is at least the denominator's. It separates a polynomial part that integrates directly from a proper rational remainder.

For a proper rational function, factor the denominator and decompose into partial fractions. Distinct linear factors receive constant numerators; repeated factors need a term for every power up to the repetition.

Verification is unusually cheap in this chapter. Differentiate the proposed antiderivative and simplify back to the original integrand.

This catches missing chain factors, sign errors in parts and incomplete partial-fraction systems. The arbitrary constant is not tested by differentiation, which is exactly why it belongs in the final family.

In this chapter

What this chapter covers

  • 01

    Antiderivatives and the arbitrary constant

  • 02

    Linearity and the reverse power rule

  • 03

    Substitution as the reverse chain rule

  • 04

    Integration by parts

  • 05

    Long division for improper rational functions

  • 06

    Partial fractions and differentiation checks

Worked example · free

Use substitution after exposing the inner derivative

Q [6 marks]. AskSia-authored practice. Find the indefinite integral of 2x/(x^2 + 4)^2 with respect to x. The marks are an AskSia study allocation, not a University marking scheme.
Both columns apply integration by parts to the product of x squared and the natural logarithm.Both columns apply integration by parts to the product of x squared and the natural logarithm. The left column differentiates the logarithm and leaves one third of x squared to integrate. The right column integrates the logarithm first, and its leftover product still contains the integral you started from.Chosen: u = ln x and v' = x^2ln x1/xdifferentiate this onex^2x^3/3integrate this oneTraded integral: one third of x^2Two short lines and it is finishedRejected: u = x^2 and v' = ln xx^22xdifferentiate this oneln xx ln x - xintegrate this oneTraded integral: 2x(x ln x - x) dxThe starting integral comes back
Both columns apply integration by parts to the product of x squared and the natural logarithm. The left column differentiates the logarithm and leaves one third of x squared to integrate. The right column integrates the logarithm first, and its leftover product still contains the integral you started from.
  • 2Choose the inner expression and rewrite its differential.
  • 2Integrate the resulting power in the new variable.
  • 2Return to x, add the constant and verify by differentiation.
Let u = x^2 + 4. Then du = 2x dx, so the entire numerator with the differential is replaced and the integral becomes the integral of u^(−2) du. Raising the exponent by one and dividing gives −u^(−1) + C. Substituting back, the antiderivative is −1/(x^2 + 4) + C. Differentiate to check: the derivative of −(x^2 + 4)^(−1) is (x^2 + 4)^(−2) times 2x, which is exactly 2x/(x^2 + 4)^2. The check also confirms that no chain factor was lost.
Sia tip — A substitution is complete only when every x and dx in the integral has been replaced by the new variable and its differential.
Glossary

Key terms

Antiderivative
An Antiderivative is a function whose derivative equals the given integrand.
Indefinite Integral
An Indefinite Integral denotes the full family of antiderivatives, including an arbitrary constant.
Substitution
Substitution changes variables to reverse a chain-rule pattern in the integrand.
Integration by Parts
Integration by Parts reverses the product rule by differentiating one factor and integrating another.
Partial Fractions
Partial Fractions decomposes a proper rational function into simpler fractions determined by the denominator factors.
FAQ

Antiderivatives and Integration Methods FAQ

Why must an indefinite integral include a constant?

Differentiation removes every constant, so infinitely many functions share the same derivative. The arbitrary constant records that entire family. Omitting it turns a general antiderivative question into one unexplained representative and loses information the original integrand cannot recover.

How do I choose between substitution and integration by parts?

Substitution is favoured when one expression is nested inside another and a multiple of its derivative appears nearby. Parts is favoured for a visible product where differentiating one factor makes it simpler. Test the proposed move mentally before committing to a page of algebra.

When should I divide before using partial fractions?

Divide whenever the numerator's degree is at least the denominator's degree. Partial fractions applies to the proper remainder, not to the original improper rational function. Skipping division produces a decomposition that cannot match the polynomial behaviour of the integrand.

Study strategy

Exam move

Sort a mixed page of integrals by method before solving any of them. Underline the inner expression for substitution, mark the factor that simplifies under differentiation for parts, and compare polynomial degrees before partial fractions. After each answer, perform a one-line derivative check; method recognition improves faster when the error is caught immediately.

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