Integrating an equation means finding the function whose derivative produces the expression you started with. If you know that the derivative of x² is 2x, then the integral of 2x is x². The process reverses differentiation, and once you learn a handful of core rules, you can integrate most equations you’ll encounter in a calculus course.
Indefinite vs. Definite Integrals
Before picking a technique, you need to know which type of integral you’re solving. An indefinite integral has no bounds and produces a function. A definite integral has upper and lower limits (like x = 1 to x = 5) and produces a number, typically representing the area under a curve between those two points.
To solve a definite integral, you first find the indefinite integral (the antiderivative), then plug in the upper bound and subtract the value at the lower bound. So every technique below applies to both types. The only difference is that final evaluation step.
The Constant of Integration
When you differentiate x² + 3, you get 2x. When you differentiate x² + 100, you also get 2x. Any constant vanishes during differentiation, which means when you reverse the process, you can’t know what constant was originally there. That’s why every indefinite integral includes “+ C” at the end, where C represents any real number. The result isn’t a single function but an infinite family of functions, all identical in shape but shifted vertically on a graph. Forgetting the + C is one of the most common mistakes in introductory calculus. Definite integrals don’t need it because the constant cancels out when you subtract the lower bound from the upper bound.
The Power Rule
The power rule handles the most common integrals you’ll see. For any term xⁿ, the integral is:
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C
You raise the exponent by one, then divide by that new exponent. So ∫ x³ dx = x⁴/4 + C, and ∫ x⁻² dx = x⁻¹/(−1) + C = −1/x + C. The rule works for any real number exponent: positive, negative, fractional, or zero. The one exception is n = −1, because dividing by zero is undefined. The integral of x⁻¹ (which is 1/x) is ln|x| + C instead.
Common Formulas Worth Memorizing
Certain integrals appear so frequently that memorizing them saves significant time. Beyond the power rule and 1/x, here are the ones you’ll use most:
- ∫ eˣ dx = eˣ + C
- ∫ cos x dx = sin x + C
- ∫ sin x dx = −cos x + C
- ∫ sec² x dx = tan x + C
- ∫ csc² x dx = −cot x + C
- ∫ sec x tan x dx = sec x + C
Each of these is just the reverse of a standard derivative. If you know your derivative table, you already know these integrals. The trig functions tan x, cot x, sec x, and csc x themselves require trig identities and substitution to integrate, so they’re handled with the techniques below rather than memorized directly.
U-Substitution
U-substitution is the first technique to reach for when the basic rules don’t immediately apply. It works whenever you can spot a function and its derivative both present in the integrand. The idea is to replace a complicated inner expression with a single variable, u, simplify, integrate, then substitute back.
Here’s the process. Say you need to integrate ∫ 18x² ⁴√(6x³ + 5) dx. You notice that 6x³ + 5 is a chunk sitting inside the fourth root, and its derivative (18x²) is right there multiplying everything. Set u = 6x³ + 5. Compute the differential: du = 18x² dx. Now replace both pieces in the original integral:
∫ 18x² (6x³ + 5)^(1/4) dx becomes ∫ u^(1/4) du
That’s just a power rule problem. Integrate to get u^(5/4) / (5/4) + C, then substitute 6x³ + 5 back in for u.
The key test: after substituting, every x in the integral (including the dx) must be gone, replaced entirely by u and du. If any x remains, either your substitution is wrong or you need a different technique. Sometimes you’ll need to solve for a leftover x term. For example, if du = −30x² dx, you can rearrange to get x² dx = −(1/30) du, then pull that fraction out front.
Integration by Parts
When you have two different types of functions multiplied together, like x·eˣ or x²·sin x, substitution usually won’t work. Integration by parts is designed for these products. The formula is:
∫ u dv = uv − ∫ v du
You split the integrand into two pieces: one you call u (which you’ll differentiate) and one you call dv (which you’ll integrate). The goal is to make the remaining integral, ∫ v du, simpler than what you started with.
Choosing which piece to call u makes or breaks this technique. The LIATE mnemonic gives you a reliable priority order. Pick u as whichever function type appears first in this list:
- L: Logarithmic functions (like ln x)
- I: Inverse trig functions (like arctan x)
- A: Algebraic functions (like x², 3x)
- T: Trig functions (like sin x, cos x)
- E: Exponential functions (like eˣ)
For ∫ x·eˣ dx, x is algebraic (A) and eˣ is exponential (E). A comes before E, so u = x and dv = eˣ dx. Then du = dx and v = eˣ. Plugging into the formula: x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C. Sometimes you need to apply integration by parts twice in a row to fully resolve the integral.
Partial Fraction Decomposition
When you’re integrating a fraction where both the top and bottom are polynomials (a rational function), partial fractions can break it into simpler pieces that are each easy to integrate on their own. One critical requirement: the degree of the numerator must be strictly less than the degree of the denominator. If it isn’t, you need to do polynomial long division first to reduce it.
The steps are straightforward once you know the setup:
- Factor the denominator as completely as possible.
- Write a partial fraction for each factor. A linear factor like (ax + b) gets a term A/(ax + b). A repeated linear factor like (ax + b)² gets two terms: A/(ax + b) + B/(ax + b)². A quadratic factor like (ax² + bx + c) gets (Ax + B)/(ax² + bx + c).
- Set the numerators equal and solve for the unknown constants (A, B, C, etc.). You can either plug in strategic x values that zero out terms, or multiply everything out and match coefficients of like powers of x.
- Integrate each simpler fraction individually. These typically produce logarithms or inverse trig functions.
For example, ∫ 1/((x + 2)(x − 3)) dx splits into A/(x + 2) + B/(x − 3). Setting x = −2 and x = 3 lets you solve for A and B quickly, and then each piece integrates as a natural log.
Choosing the Right Technique
With several methods available, a simple decision process helps. Start by checking whether the integral matches a basic formula or power rule directly. If not, look for a composite function where the inner function’s derivative appears elsewhere in the expression, which signals u-substitution. If you see two different function types multiplied together (especially involving ln x, inverse trig, or a polynomial times an exponential or trig function), try integration by parts. If you’re dealing with a ratio of polynomials, set up partial fractions. Many integrals require combining techniques: you might use substitution inside an integration by parts problem, or apply partial fractions after a trig substitution.
You can always verify your answer by differentiating it. If the derivative of your result gives back the original integrand, you integrated correctly. This check takes seconds and catches most errors.
When No Formula Exists
Some functions simply don’t have antiderivatives that can be written with standard formulas. The classic example is e^(−x²), which appears constantly in statistics and physics. For definite integrals of these functions, numerical methods give you an approximate answer. The trapezoidal rule estimates the area under a curve by dividing it into trapezoids. Simpson’s rule uses parabolic arcs instead and converges to the true answer much faster, with its error shrinking proportionally to 1/n⁴ (where n is the number of subdivisions) compared to 1/n² for the trapezoidal rule. In practice, a calculator or computer handles these computations, but understanding the concept explains why some integrals have exact answers and others are approximated.
Why Integration Matters Beyond Math Class
Integration has direct physical meaning. Whenever a quantity accumulates over time or space, integration is the tool that computes the total. Integrating a velocity function gives total displacement. Integrating a force over a distance gives the work done, which is how engineers calculate the energy needed to compress a spring or pump water out of a tank. In manufacturing, integration computes the volume of curved solids by rotating a cross-section around an axis. Density functions that vary across an object get integrated to find total mass. Hydrostatic force on a submerged surface, the probability under a bell curve, the total charge in a circuit: all computed by integration. The technique you’re learning isn’t just an academic exercise. It’s the mathematical backbone of nearly every quantitative field.