Basic Integration Techniques

⚡ Key Takeaways

Integration reverses differentiation: given a function's rate of change, it recovers the original function (up to a constant). The basic techniques are a toolkit — direct power rule, substitution, integration by parts, partial fractions, and trigonometric methods — for finding antiderivatives and computing the areas they represent. Used everywhere area, accumulation, or total-from-rate appears. Foundations built by Newton & Leibniz (1660s–1680s), formalized through the 18th–19th centuries.

💡 The Core Insight

Every integration technique is the reverse of a differentiation rule. The power rule run backward gives the power rule for integrals. The chain rule run backward gives substitution. The product rule run backward gives integration by parts. If you can recognize which differentiation rule would have produced your integrand, you can undo it. Integration is pattern-matching against the derivative rules you already know.

DefinitionAntiderivative and Indefinite Integral

A function $F$ is an antiderivative of $f$ on an interval if $F'(x) = f(x)$ for all $x$ in that interval. The indefinite integral is the family of all antiderivatives:

1

\quad \int f(x)\,dx = F(x) + C, \qquad \text{where } F'(x) = f(x)

Derivation hint: If $F' = f$ and $G' = f$ , then $(F-G)' = 0$ , so $F - G$ is constant — hence the arbitrary constant $C$ . Here $f(x)$ is the integrand, $x$ is the variable of integration, $dx$ marks that variable, and $C \in \mathbb{R}$ is the constant of integration.

DefinitionDefinite Integral

For a continuous function $f$ on $[a,b]$ , the definite integral is the signed area under the curve, defined as the limit of Riemann sums:

2

\quad \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x, \qquad \Delta x = \frac{b-a}{n}

Derivation hint: Partition $[a,b]$ into $n$ strips of width $\Delta x$ , approximate each by a rectangle of height $f(x_i^*)$ , sum, and let $n\to\infty$ .

TheoremFundamental Theorem of Calculus

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ , then the definite integral equals the change in the antiderivative:

3

\quad \int_a^b f(x)\,dx = F(b) - F(a) = \Big[F(x)\Big]_a^b

Derivation hint: Define $G(x)=\int_a^x f(t)\,dt$ ; show $G'(x)=f(x)$ (FTC Part 1), so $G$ and $F$ differ by a constant, which cancels in $F(b)-F(a)$ . This is the bridge that turns the area problem [2] into the antiderivative problem [1] — making integration practical.

DefinitionThe Five Basic Techniques

1. Power Rule (reverse):

4

\quad \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \ne -1), \qquad \int x^{-1}\,dx = \ln|x| + C

Derivation hint: Differentiate $\dfrac{x^{n+1}}{n+1}$ to recover $x^n$ ; the case $n=-1$ would divide by zero, so it needs $\ln|x|$ , whose derivative is $1/x$ .

2. Substitution (reverse chain rule): with $u = g(x)$ , $du = g'(x)\,dx$ ,

5

\quad \int f(g(x))\,g'(x)\,dx = \int f(u)\,du = F(u) + C

Derivation hint: The chain rule gives $\frac{d}{dx}F(g(x)) = F'(g(x))g'(x) = f(g(x))g'(x)$ ; integrating both sides reverses it.

3. Integration by Parts (reverse product rule):

6

\quad \int u\,dv = uv - \int v\,du

Derivation hint: Integrate the product rule $(uv)' = u'v + uv'$ , rearrange to isolate $\int uv'\,dx = \int u\,dv$ .

4. Partial Fractions: decompose a proper rational function into simpler pieces,

7

\quad \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}, \quad \text{then integrate each as } \ln

Derivation hint: Multiply through by the denominator and match coefficients (or substitute roots) to solve for $A,B$ .

5. Trigonometric Standard Forms:

8

\quad \int \sin x\,dx = -\cos x + C, \qquad \int \cos x\,dx = \sin x + C, \qquad \int \sec^2 x\,dx = \tan x + C

Derivation hint: Each is the reverse of a known trig derivative — e.g. $\frac{d}{dx}(-\cos x)=\sin x$ .

ProofIntegration by Parts from the Product Rule

We derive [6], the workhorse of the toolkit.

Step 1. Start from the product rule for two differentiable functions $u(x)$ and $v(x)$ :

\frac{d}{dx}\big(u\,v\big) = u'\,v + u\,v'.

Step 2. Integrate both sides with respect to $x$ . The left side integrates trivially (it is an exact derivative):

u\,v = \int u'\,v\,dx + \int u\,v'\,dx.

Step 3. Rearrange to isolate one of the integrals:

\int u\,v'\,dx = u\,v - \int u'\,v\,dx.

Step 4. Write in differential form using $dv = v'\,dx$ and $du = u'\,dx$ :

\int u\,dv = uv - \int v\,du. \qquad \blacksquare

The method trades a hard integral $\int u\,dv$ for a (hopefully easier) one $\int v\,du$ . Success depends on choosing $u$ to become simpler when differentiated.

∎

🧠 The LIATE Choosing Rule

For integration by parts, which factor should be $u$ ? The mnemonic LIATE ranks candidates by how nicely they differentiate: Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. Pick $u$ as the function highest on the list — it will simplify (or vanish) under repeated differentiation — and let the rest be $dv$ . For $\int x\,e^x\,dx$ , $x$ (Algebraic) beats $e^x$ (Exponential), so $u=x$ , $dv=e^x dx$ .

IN

Isaac Newton

1643–1727

Newton developed his "method of fluxions" (calculus) around 1665–1666 during the plague years at Woolsthorpe. He treated integration as finding the "fluent" quantity whose rate of change ("fluxion") was given — exactly the antiderivative idea. His geometric and physical motivation (areas, velocities, gravitation) drove the subject, though his notation $\dot{x}$ lost out to Leibniz's $\int$ and $dx$ , which dominate today.

🧠 Substitution as Re-Labeling

Substitution works because integration doesn't care what we call the variable — only that the $du$ matches the rest of the integrand. When you see a function and its derivative both present, you can "zoom in" on the inner function $g(x)$ , rename it $u$ , and the messy chain-rule structure collapses into a clean integral in $u$ . Think of it as undoing a disguise: the chain rule dressed up a simple integral, and substitution strips the costume off.

Figure

Substitution geometry: the area under

f(g(x))g'(x)

in

x

equals the area under

f(u)

in

u

— the

du

factor stretches/compresses the horizontal axis exactly right.

ExampleDefinite Integral by Substitution

Compute $\displaystyle\int_0^2 x\,e^{x^2}\,dx$ .

Given: integrand $x\,e^{x^2}$ , limits $0$ to $2$ .

Step 1 — Spot the inner function. Let $u = x^2$ . Then $du = 2x\,dx$ , so $x\,dx = \tfrac{1}{2}\,du$ .

Step 2 — Change the limits (so we never convert back):

when $x=0$ : $u=0$
when $x=2$ : $u=4$

Step 3 — Rewrite:

\int_0^2 x\,e^{x^2}\,dx = \int_0^4 e^u \cdot \tfrac{1}{2}\,du = \tfrac{1}{2}\int_0^4 e^u\,du.

Step 4 — Integrate and evaluate:

\tfrac{1}{2}\big[e^u\big]_0^4 = \tfrac{1}{2}(e^4 - e^0) = \tfrac{1}{2}(e^4 - 1).

\boxed{\displaystyle\int_0^2 x\,e^{x^2}\,dx = \tfrac{1}{2}(e^4 - 1) \approx 26.80}

(dimensionless — a pure number representing area in $x$ – $y$ units).

ExampleIntegration by Parts

Compute $\displaystyle\int x\ln x\,dx$ .

Given: product of algebraic $x$ and logarithmic $\ln x$ .

Step 1 — Choose $u$ via LIATE. Logarithm ranks above Algebraic, so:

u = \ln x, \qquad dv = x\,dx.

Step 2 — Differentiate and integrate:

du = \frac{1}{x}\,dx, \qquad v = \frac{x^2}{2}.

Step 3 — Apply [6], $\int u\,dv = uv - \int v\,du$ :

\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac{1}{x}\,dx = \frac{x^2}{2}\ln x - \frac{1}{2}\int x\,dx.

Step 4 — Finish the remaining integral:

= \frac{x^2}{2}\ln x - \frac{1}{2}\cdot\frac{x^2}{2} + C.

\boxed{\displaystyle\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C}

Check: differentiate — $x\ln x + \tfrac{x}{2} - \tfrac{x}{2} = x\ln x$ . ✓

⚠️ The Most Frequent Errors

1. Forgetting $+C$ . Every indefinite integral has an arbitrary constant. Omitting it loses an entire family of solutions and breaks differential-equation problems. 2. $\int \frac{1}{x}\,dx \ne \frac{x^{0}}{0}$ . The power rule [4] explicitly excludes $n=-1$ . The answer is $\ln|x| + C$ . 3. No product/quotient rule for integrals. There is no formula $\int fg = (\int f)(\int g)$ . You must use substitution or by parts. 4. Forgetting to change limits in definite substitution (or to convert $u$ back to $x$ if you don't). Mixing $u$ -limits with an $x$ -expression gives a wrong number. 5. Dropping the $dx$ . It is not decoration — in substitution the $dx$ becomes $du$ via the derivative, and the bookkeeping must balance.

CounterexampleWhy

\int fg \ne (\int f)(\int g)

Take $f(x) = g(x) = x$ . Then $\int x\cdot x\,dx = \int x^2\,dx = \dfrac{x^3}{3} + C$ . But the "product of integrals" would give $\left(\dfrac{x^2}{2}\right)\left(\dfrac{x^2}{2}\right) = \dfrac{x^4}{4}$ — a completely different function. These disagree even in their degree, proving no such product rule exists.

connection · Where These Techniques Sit

Substitution is literally the Chain Rule read backward, and integration by parts is the Product Rule reversed — so mastering Differentiation is mastering half of integration. The definite integral connects to the Fundamental Theorem of Calculus, which links it to areas and to Riemann Sums. Partial fractions relies on Polynomial Factorization and the algebra of Rational Functions. These methods are the entry point to Differential Equations, where integration recovers a function from its rate of change, and to Improper Integrals when limits become infinite.

🔬 Where Integration Techniques Are Used

⏳ History

- c. 250 BCE — Archimedes computes the area of a parabolic segment by the "method of exhaustion," a Riemann-sum ancestor.

- 1635 — Cavalieri publishes his "method of indivisibles," treating areas as sums of infinitely thin lines.

- 1650s — Fermat and Wallis find

\int x^n\,dx

for rational

n

, anticipating the power rule.

- 1665–1666 — Newton develops fluxions, relating areas to antiderivatives during his Woolsthorpe years.

- 1675 — Leibniz introduces the integral sign

\int

(an elongated "S" for summa) and the

dx

notation still used today.

- 1690s — The Bernoullis (Jacob and Johann) and Euler systematize substitution and develop integration by parts.

- 1768 — Euler's Institutiones calculi integralis compiles integration techniques into a comprehensive treatise.

- 1820s — Cauchy gives the first rigorous limit-based definition of the definite integral.

- 1854 — Riemann formalizes the integral as a limit of sums [2], clarifying exactly which functions are integrable.

open-problem · The Boundary of the Elementary

Differentiation always succeeds: any combination of elementary functions has an elementary derivative. Integration does not. Liouville (1830s) and later Risch (1968) proved that some elementary functions have no elementary antiderivative — for example $\int e^{-x^2}\,dx$ , $\int \frac{\sin x}{x}\,dx$ , and $\int \frac{1}{\ln x}\,dx$ . The Risch algorithm can decide whether a given elementary integrand has an elementary antiderivative, but it is enormously complex and not fully implemented in any computer algebra system. This means the "basic techniques" form a complete toolkit only for a subset of integrals; beyond it lie special functions (the error function $\mathrm{erf}$ , the logarithmic integral $\mathrm{li}$ ) and numerical methods. Knowing when to stop searching for a closed form is itself a skill.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also