Integration is one half of calculus, and the more mysterious half to most students, because where differentiation has a clear mechanical feel, integration often looks like guesswork. It is not. Once you see that integration simply undoes differentiation, and once you have a small set of standard results and a clear method for choosing between techniques, almost every integral you meet becomes a matter of recognising which tool applies. This guide builds that toolkit from the ground up: antiderivatives, the standard integrals, the rules that combine them, definite integrals and the areas they measure, two economic applications, and finally the three big techniques, substitution, integration by parts, and partial fractions.
Antiderivatives and the Indefinite Integral
Integration starts from a simple reversal. If a function F(x) has derivative f(x), then F(x) is called an antiderivative of f(x).
The catch is that antiderivatives are never unique, because adding any constant leaves the derivative unchanged.
To see this, differentiate both 4x squared and 4x squared plus 7.
Both give 8x, because the constant 7 has zero derivative, which is precisely why a function has a whole family of antiderivatives rather than just one.
The indefinite integral packages that entire family into a single expression by tacking on an arbitrary constant of integration.
It is called indefinite because c is unknown. The function being integrated, f(x), is the integrand; F(x) is an antiderivative; and c absorbs all the others. So integrating 8x means asking what differentiates to 8x, which is 4x squared, and writing the result with its constant.
The Standard Integrals
A handful of results, each the reverse of a known derivative, cover most integrands. The power rule comes from reversing the power rule for differentiation: raise the power by one and divide by the new power.
The exclusion of n equals minus one matters, because dividing by zero is undefined, and that single case has its own result.
The absolute value is essential, since the natural logarithm is only defined for positive inputs while x itself may be negative. As a worked case, take n equal to minus three halves, raise it by one to minus one half, and divide.
The exponential and logarithmic standards are equally compact. The exponential function is its own integral, the general exponential picks up a logarithm in the denominator, and the integral of the natural log is a result we will derive later by parts.
So integrating 3 to the x just applies the general exponential rule with base 3.
The two basic trigonometric integrals follow from reversing the derivatives of sine and cosine, and the minus sign on the first is the part students most often drop.
The Linear Combination Rule
Integration distributes over sums, differences, and constant multiples, exactly as differentiation does, which is what lets you break a long integrand into manageable pieces.
Applying it to a mixed integrand, you split first and then integrate each piece with its standard rule.
Definite Integrals
A definite integral carries limits a and b, and its result is a number rather than a family of functions. You find any antiderivative, then subtract its value at the lower limit from its value at the upper limit.
The constant of integration cancels in the subtraction, so it is simply omitted. Evaluating a straightforward case shows the rhythm: integrate, bracket with the limits, then substitute and subtract.
The linear combination rule carries over unchanged; the limits simply travel with each sub-integral.
Area Under a Curve
The definite integral’s most famous meaning is area, but you have to be careful about sign. When the integrand is non-negative across the interval, the integral is exactly the area between the curve, the x-axis, and the two vertical boundaries.
For the curve 4 minus x squared between minus one and one, which stays positive throughout, the area comes out cleanly.
When the curve lies below the axis, the integral is negative, and the area is its magnitude.
For 4 minus 2x between two and four, the integral evaluates to minus four, so the area is four, which a quick check against the triangle formula confirms.
The subtle case is when the curve crosses the axis inside the interval. There, integrating straight from a to b gives the net signed area, with the parts below cancelling the parts above, not the total area. You must find the crossing points, split the integral at each, and add the magnitudes of the pieces. For 1 minus x squared from minus two to two, the curve crosses at x equals plus and minus one, giving three pieces whose magnitudes are each four thirds.
The direct integral over the whole interval gives minus four thirds, which is emphatically not the area, the clearest reminder that signed area and total area are different things.
Two Economic Applications
Integration recovers a total from a rate, which is exactly the relationship between marginal cost and cost. Since marginal cost is the derivative of the cost function, integrating it recovers the cost, and the constant is pinned down by the fixed costs, the cost when output is zero.
For a marginal cost of 2q plus 100 times e to the q, with fixed costs of ten thousand, integrating gives q squared plus 100 e to the q plus a constant, and the condition that cost at zero equals ten thousand fixes that constant at 9,900.
The second application is consumer and producer surplus. At a market equilibrium with quantity q-star and price p-star, consumers collectively pay less than they would have been willing to, and producers receive more than they strictly needed; those gaps are the two surpluses, each an area found by integration.
With demand 70 minus a third of q and supply 20 plus half of q, setting them equal gives an equilibrium quantity of 60 and price of 50, and the integrals then yield a consumer surplus of 600 and a producer surplus of 900.
Integration by Substitution
Substitution reverses the chain rule, and it is the technique for integrands built from a function of an inner function. The simplest case is a linear inner function ax plus b, where the substitution always leaves behind a tidy factor of one over a.
The method itself is three moves: set g equal to the inner function, express dx in terms of dg, then integrate and back-substitute. For e to the 3x plus 1, set g equal to 3x plus 1 so that dx is one third dg.
When the inner function is non-linear, substitution only works if the integrand also contains a factor matching the derivative of the inner function, up to a constant. That factor is what funds the switch from dx to dg. For x over x squared plus one, setting g equal to x squared plus one gives dg equal to 2x dx, and the x in the numerator supplies exactly the factor needed.
The same idea handles trigonometric compositions. The integral of tan x looks unpromising until you write it as sine over cosine and set g equal to cosine, whose derivative is minus sine, which the numerator provides.
Integration by Parts
Where substitution reverses the chain rule, integration by parts reverses the product rule. It rewrites the integral of a product as a boundary term minus a new integral that is, with luck, simpler.
The whole skill is choosing which factor is f and which is g prime. Pick f to be the factor that gets simpler when differentiated, and g prime to be the factor you can integrate easily; a reliable priority for f is logarithm, then polynomial, then exponential. For x times ln x, the log simplifies on differentiation, so f is ln x and g prime is x.
This is also the technique behind the earlier result for the integral of ln x, taking f as ln x and g prime as one.
Partial Fractions
A rational integrand, one polynomial over another, is often impossible to integrate as written but easy once split into simpler fractions, provided the numerator’s degree is below the denominator’s. The form of the split depends on the denominator’s factors.
The first case is distinct linear factors, where each factor contributes one simple fraction that integrates to a logarithm.
For x over x squared minus x minus two, factoring the denominator into (x minus 2)(x plus 1) and solving for the constants gives two thirds and one third.
The second case is a repeated linear factor. A root repeated m times contributes m terms, one for each power from one up to m.
For (x plus 3) over (x plus 2)(x minus 1) squared, the repeated root produces a term in (x minus 1) and another in its square, and the squared term integrates to a reciprocal rather than a logarithm.
The third case is an irreducible quadratic, one with negative discriminant that cannot be factored over the reals. Its term carries a linear numerator, and after completing the square it integrates to a combination of a logarithm and an inverse tangent.
For x over (x minus 1)(x squared plus 2x plus 2), where the quadratic has discriminant minus four, the decomposition and a completing-the-square step lead to all three function types in the answer.
Definite Integrals With the Two Techniques
Both substitution and parts extend to definite integrals, with one adjustment each. For substitution, rather than back-substituting at the end, you convert the limits from x-values to g-values as you go.
For x times e to the (x squared plus 1) from zero to one, setting g equal to x squared plus one turns the limits into one and two, and the integral resolves without ever returning to x.
For integration by parts, the boundary term is now evaluated at the limits rather than carried as a function.
For x times e to the x from zero to one, with f equal to x and g prime equal to e to the x, the boundary term gives e and the remaining integral gives e minus one.
Choosing the Right Method
The hardest part of integration in practice is not executing a technique but recognising which one to use, and a short checklist settles most cases. First, is the integrand a single standard form, a power, an exponential, a reciprocal, a sine or cosine? Then integrate it directly. Is it a linear combination of standard forms? Split it and integrate term by term. Is it a standard form wrapped around a linear inner function ax plus b? Substitute, and expect a factor of one over a. Is it a product in which one factor is the derivative of the inner part of the other, up to a constant? Substitute with g as that inner function. Is it a product of two unrelated functions, such as a polynomial times a logarithm or a polynomial times an exponential? Use integration by parts, choosing f as the part that simplifies on differentiation. And is it a rational function with numerator degree below denominator degree? Use partial fractions, factoring the denominator and applying the case that matches.
For quick recall, the standard integrals are these.
Practice exercises here: https://datalad.co.uk/integration-workbook-10-exercises-with-full-solutions/
Conclusion
Integration is the reverse of differentiation, and everything in this guide follows from that one idea. Learn the standard integrals cold, since they are the vocabulary of the subject, and use the linear combination rule to break complex integrands into them. Treat the indefinite integral as a whole family with its constant c, and the definite integral as a single number found by subtracting the antiderivative’s values at the limits, taking care that area means the magnitude, split at every axis crossing. When an integrand is not a standard form, run the checklist: substitution for compositions, parts for products of unlike functions, partial fractions for rational functions. Verify by differentiating whenever you are unsure, because the answer to “did I integrate correctly” is always just one derivative away.
See you soon.
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