Method · Calculus
How to Integrate: Step by Step
Integration is mostly pattern recognition — matching the shape in front of you to a rule for reversing it, and always checking the result by differentiating back.
Read the shape first
Learning how to integrate is mostly learning to recognise which reverse-rule an expression calls for. Every rule below is a differentiation rule read backward, so the shape of the integrand — not a fixed formula — decides which one applies.
| The integrand is | Shape | Rule |
|---|---|---|
| a sum or difference | u + v | integrate term by term |
| a constant times something | k · u | the constant stays, integrate u |
| a power of x | xⁿ | power rule for integration |
| a standard function of a linear expression | sin(ax+b), e^(ax+b) | standard form, divide by a |
| 1 over a linear expression | 1/(ax+b) | logarithm case |
| a genuine product of two functions of x | u · v | integration by parts (beyond this guide) |
The last row matters as much as the others: recognising that a shape needs a technique you have not covered yet is not a failure, it is the correct diagnosis.
The power rule, reversed
Differentiating xⁿ multiplies by n and lowers the exponent by one. Integrating does the opposite in the opposite order: raise the exponent by one, then divide by the new exponent.
- ∫ 3x² + 5x dxa sum, so integrate term by term
- 3x³/3 + 5x²/2raise each exponent by one, divide by the new exponent
- x³ + 5x²/2 + Cthe coefficient 3 cancels against the division by 3
The rule works for negative and fractional exponents too, with one exception. x⁻² integrates to x⁻¹/(−1), which is −1/x. But x⁻¹ — that is, 1/x — cannot use this rule at all, because raising the exponent by one would mean dividing by zero.
The exception: 1/x integrates to a logarithm
When the power rule would divide by zero, a different antiderivative takes over: 1/x integrates to ln|x|. The absolute value matters, because ln is undefined for negative inputs but 1/x is perfectly well defined there.
- ∫ 1/(2x + 1) dxthe reciprocal of a linear expression
- ln|2x + 1|the logarithm case
- ln|2x + 1| / 2 + Cdivide by the coefficient of x, as in linear substitution
Standard functions, and linear substitution
Sin, cos, eˣ and the rest each have a fixed antiderivative you learn once — sin integrates to −cos, cos integrates to sin, eˣ integrates to itself. These standard forms are worth memorising outright rather than re-deriving each time. The complication is what happens when the input is not bare x but a linear expression like 2x + 1.
- ∫ sin(2x) dxsin of a linear expression
- −cos(2x)the ordinary antiderivative of sin, applied to the inside
- −cos(2x) / 2 + Cdivide by the coefficient of x — this undoes the chain rule
What integration cannot shortcut: products and non-linear insides
Two shapes need more than the rules above. A genuine product of two functions of x, like x·sin(x), needs integration by parts — there is no way to integrate term by term when the terms are multiplied rather than added. A non-linear expression inside a standard function, like sin(x²), needs a proper u-substitution, not the constant division that works for a linear inside.
A u-substitution replaces part of the expression with a new variable u chosen so that the rest of the integral, including dx, rewrites cleanly in terms of u and du. For a linear inside like sin(2x), setting u = 2x makes du = 2 dx, and dividing by the constant 2 is really this substitution done in one step without the extra notation. For sin(x²), setting u = x² gives du = 2x dx — and unless a matching 2x already sits outside the sine, that substitution does not clean up the integral, which is exactly why sin(x²) resists the shortcuts above.
Recognising these two shapes before attempting a shortcut that does not exist is the difference between a wrong answer and an honest 'this needs a technique I have not used yet'.
Indefinite integrals keep the +C, definite integrals lose it
Everything above, without limits attached, is an indefinite integral: a family of functions ending in +C, the constant of integration that stands in for whichever constant differentiation destroyed. Add a lower and upper limit and the question changes — a definite integral evaluates to one specific number, and that number does not depend on which member of the family you picked.
This is the fundamental theorem of calculus: pick any antiderivative F, evaluate it at the upper limit, subtract its value at the lower limit, and the constant of integration cancels out of the subtraction no matter what it was.
Definite integrals: evaluate, then subtract
Once an antiderivative F is found, a definite integral from a to b is F(b) − F(a). The +C is never needed here, because it appears in both terms and cancels.
- F(x) = x³/3any antiderivative — the +C is dropped
- F(1) − F(0) = ⅓ − 0evaluate at the upper limit, then the lower
- ⅓subtract
Checking your answer, always
Differentiate what you found. If it reproduces the original expression exactly, the integration is correct — this is the one check that never fails to catch a mistake, because differentiation is a mechanical procedure with no ambiguity. Make it a habit to check by differentiating every antiderivative before trusting it, the same way you would check a factoring answer by expanding it back out.
The Integral Calculator on this site runs exactly this check internally: it differentiates its own answer and compares it numerically to the original expression before showing anything. If the two do not match, it says the result could not be verified rather than showing a possibly wrong antiderivative.
Questions about how to integrate
What is the first step in integrating an expression?
Identify the shape: a sum integrates term by term, a constant multiplier stays put, and everything else is matched against the power rule, a standard form, or flagged as needing a technique like integration by parts.
Why divide by the coefficient when the inside is linear?
Because differentiating the result would otherwise bring back that coefficient via the chain rule. Dividing by it in advance cancels that factor, so the check — differentiate the answer, compare to the original — comes out exact.
Why does 1/x integrate to a logarithm instead of using the power rule?
The power rule for xⁿ divides by the new exponent, n + 1. For 1/x that would mean dividing by zero, so a separate rule applies: 1/x integrates to ln|x|.
How do I know if an expression needs integration by parts?
When it is a genuine product of two functions of x — like x·sin(x) or x²·eˣ — and neither factor is a plain constant. A constant times a function of x is not a product in this sense; only the constant multiple rule is needed there.