Integral calculus gives you five core strategies for finding antiderivatives: u-substitution, integration by parts, trig integrals, trig substitution, and partial fractions. Picking the right one depends on the shape of what you are integrating. This document covers the first two strategies in depth, with worked examples and decision cues.
Difficulty: Intermediate | Prerequisites: Basic differentiation rules, chain rule, product rule, and familiarity with common antiderivatives.
Most integrals you meet in a calculus course do not yield to a single formula. This unit teaches you a decision tree: look at the integrand, identify its structure, and route it to the technique that simplifies it. U-substitution reverses the chain rule. Integration by parts reverses the product rule. The remaining three techniques (trig integrals, trig substitution, partial fractions) handle families of integrands that neither of the first two can crack alone. Mastering the selection process matters as much as mastering each technique, because exams test both.
Antiderivative
A function F(x) whose derivative equals f(x). In simple terms, it is the reverse of differentiation.
Indefinite integral
The family of all antiderivatives of f(x), written ∫ f(x) dx = F(x) + C. Think of it as "undo the derivative, then add an arbitrary constant."
U-substitution (change of variable)
A technique where you set u = g(x), compute du = g'(x) dx, and rewrite the integral entirely in terms of u. In simple terms, you spot a function and its derivative sitting together inside the integral, and you collapse them into a simpler variable.
Integration by parts
A technique based on the product rule in reverse: ∫ u dv = uv - ∫ v du. Think of it as choosing which factor to differentiate and which to integrate, then trading one integral for a (hopefully) simpler one.
Cyclic integral
An integral where applying integration by parts eventually reproduces the original integral, allowing you to solve for it algebraically. In simple terms, the integral "comes back around" and you treat it as an unknown to isolate.
Integrand
The expression being integrated, the bit between ∫ and dx.
Look for a function and its derivative both appearing in the integrand. If you can write the integral as ∫ f(g(x)) · g'(x) dx, set u = g(x).
Identify an inner function g(x) and check that g'(x) (or a constant multiple of it) appears as a factor.
Set u = g(x), compute du = g'(x) dx.
Rewrite the entire integral in terms of u and du, with no leftover x.
Integrate in u, then substitute back.
Set u = ln x, so du = (1/x) dx.
The integral becomes ∫ u^(-1/2) du = 2u^(1/2) + C = 2√(ln x) + C.
Set u = 4 + 3x, so du = 3 dx, meaning (1/3) du = dx.
Change limits: when x = 0, u = 4; when x = 1, u = 7.
I = (1/3) ∫₄⁷ u^(-1/2) du = (1/3) · [2u^(1/2)]₄⁷ = (2/3)(√7 - √4) = (2/3)(√7 - 2).
If after substitution you still have x terms that will not convert to u, u-sub alone will not work. Move on to another technique.
The integrand is a product of two functions where one simplifies when differentiated and the other is easy to integrate. Also use it when the integral becomes cyclic (reproduces itself after one or two applications).
∫ u dv = uv - ∫ v du
Choose u as the factor that becomes simpler when you differentiate it. Choose dv as the rest, which you must be able to integrate.
Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential.
Pick u from the category that appears earliest in this list. The idea is that logs and inverse trig functions are hard to integrate but easy to differentiate, so they should be u.
First, use the substitution u = √x, so x = u², dx = 2u du.
I = ∫ eᵘ · 2u du = 2 ∫ u eᵘ du.
Now apply integration by parts with u = u, dv = eᵘ du, so du = du, v = eᵘ.
I = 2[u eᵘ - ∫ eᵘ du] = 2(u eᵘ - eᵘ) + C = 2(√x · e^(√x) - e^(√x)) + C.
Sometimes two rounds of integration by parts bring back the original integral. When that happens, call the original integral I, set up the equation, and solve for I algebraically. This commonly occurs with products of exponentials and trig functions, such as ∫ eˣ sin x dx.
U-substitution formula
∫ f(g(x)) · g'(x) dx = ∫ f(u) du, where u = g(x)
Integration by parts formula
∫ u dv = uv - ∫ v du
Definite-integral version (u-sub)
When using u-sub on a definite integral, convert the limits along with the variable. If u = g(x), then ∫ₐᵇ f(g(x)) g'(x) dx = ∫_{g(a)}^{g(b)} f(u) du. No back-substitution is needed.
U-substitution and integration by parts underpin nearly every applied integral in physics and engineering. Calculating the work done by a variable force, the charge accumulated in a circuit over time, or the probability under a continuous distribution all rely on these techniques. If you go on to differential equations, integration by parts is the backbone of solving many standard forms.
Students often forget to convert dx when doing u-sub. Setting u = g(x) is only half the job; you must also replace dx with du/g'(x). Leaving a stray dx in terms of x is the most common error.
When using integration by parts, students sometimes choose u and dv so that the new integral ∫ v du is harder than the original. If that happens, swap your choices.
Students assume u-sub will always work when they spot a composite function. It only works when the derivative of the inner function (up to a constant) also appears as a factor.
In definite integrals with u-sub, students frequently forget to change the limits of integration and then try to plug the original x-limits into a function of u. Either change the limits or substitute back to x before evaluating.
⚠️ U-sub is typically the first technique tested. Exam questions often embed a clean substitution inside a messier-looking integral to see whether you can spot it.
⚠️ Integration by parts questions commonly pair a logarithm or inverse trig function with a polynomial. Know the LIATE mnemonic.
⚠️ Expect at least one problem that combines u-sub with another technique (e.g., u-sub first, then integration by parts on the simplified integral, as in the ∫ e^(√x) dx example).
⚠️ Definite integrals with u-sub: markers penalise wrong limits. Always convert or always back-substitute; do not mix the two approaches.
True or false: u-substitution requires the derivative of the inner function to appear as a factor in the integrand.
Fill in the blank: The integration by parts formula is ∫ u dv = _______.
True or false: When applying u-sub to a definite integral, you must always substitute back to x before evaluating.
Fill in the blank: In the LIATE rule, the L stands for _______.
True or false: If two applications of integration by parts reproduce the original integral, you should start over with a different method.
Answers: 1. True (up to a constant multiple). 2. uv - ∫ v du. 3. False; you can convert the limits instead. 4. Logarithmic. 5. False; set the original integral equal to I and solve algebraically.
Q: Evaluate ∫ dx / (x ln x).
A: Set u = ln x, du = (1/x) dx. The integral becomes ∫ u^(-1/2) du = 2√u + C = 2√(ln x) + C.
Q: Evaluate ∫ e^(√x) dx.
A: First substitute u = √x, so dx = 2u du. The integral becomes 2 ∫ u eᵘ du. Apply integration by parts with u = u, dv = eᵘ du to get 2(u eᵘ - eᵘ) + C = 2(√x · e^(√x) - e^(√x)) + C.
Q: Evaluate ∫₀¹ 1/√(4 + 3x) dx.
A: Set u = 4 + 3x, du = 3 dx. Limits become u = 4 to u = 7. I = (1/3) ∫₄⁷ u^(-1/2) du = (2/3)(√7 - 2).
Q: When should you use integration by parts rather than u-substitution?
A: Use integration by parts when the integrand is a product of two different types of function (e.g., polynomial times exponential, logarithm times polynomial) and you cannot identify a clean u-sub. The LIATE rule helps you assign u and dv.
Q: What is a cyclic integral and how do you solve one?
A: A cyclic integral is one where repeated integration by parts reproduces the original integral. Label the original integral I, carry through the algebra, and then solve the resulting equation for I.
U-substitution is the integral counterpart of the chain rule from differential calculus. If you are comfortable differentiating composite functions, you already have the intuition for spotting u-sub candidates.
Integration by parts connects to the product rule and recurs heavily in differential equations (e.g., solving ∫ x eˣ dx arises in first-order linear ODEs). It also underpins the derivation of reduction formulas for powers of trig functions, which you will meet in the trig integrals unit.
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