Evaluating Integrals

Direct Antidifferentiation

Recognizing Standard Forms

Setting Up Definite Integrals

Handling Absolute Values

Summary: The Evaluation Strategy Ladder

Evaluating Integrals FAQ

Putting It All Together

Evaluating integrals draws on everything: recognizing standard forms, selecting appropriate techniques, handling special cases, and verifying results. The process is part pattern recognition, part strategic choice, part careful execution.

Start simple. Direct antidifferentiation works more often than expected—many integrals match known formulas or yield to basic algebra. When direct methods fail, systematically consider substitution, parts, and other techniques.

For definite integrals, setup matters as much as computation. Identify the correct bounds, express the integrand properly, and watch for discontinuities that signal improper integrals. A well-posed integral is half solved.

Key Terms

Definite Integral— correct setup: bounds, integrand, variable

Indefinite Integral— always includes

+ C

Antiderivative— differentiate to verify:

F'(x)

should return

f(x)

Integrand— simplify before reaching for techniques

Bounds of Integration— convert when substituting in definite integrals

See All Calculus Definitions →

Direct Antidifferentiation

Try the straightforward approach first. Does the integrand match a standard form?

Direct Antidifferentiation

\int (3x^2 + 5e^x - \sec^2 x)\, dx

Apply linearity and known formulas:

Direct Antidifferentiation

= 3 \cdot \frac{x^3}{3} + 5e^x - \tan x + C = x^3 + 5e^x - \tan x + C

Algebraic simplification often reveals standard forms:

Direct Antidifferentiation

\int \frac{x^3 + 1}{x}\, dx = \int \left(x^2 + \frac{1}{x}\right)\, dx = \frac{x^3}{3} + \ln|x| + C

Pitfall: Integrating the top and bottom separately

Integrating

x^3 + 1

and

x

on their own gives

\frac{x^4/4 + x}{x^2/2} = \frac{x^2}{2} + \frac{2}{x}

, whose derivative at

x = 1

-1

, while the integrand there is

2

. Splitting the fraction first gives

\frac{x^3}{3} + \ln|x| + C

Expand, simplify, and separate before reaching for techniques.

Here the denominator is a single term, so the fraction breaks into one fraction per term of the numerator:

\frac{x^3 + 1}{x} = \frac{x^3}{x} + \frac{1}{x}

. Each piece is then a power of

x

, and the two standard forms

\int x^n\, dx

and

\int \frac{1}{x}\, dx

finish the job. Linearity does the rest, one term at a time.

Recognizing Standard Forms

Many integrals are disguised versions of basic formulas.

Completing the square transforms quadratics:

Recognizing Standard Forms

\int \frac{1}{x^2 + 4x + 8}\, dx = \int \frac{1}{(x+2)^2 + 4}\, dx

This matches

\int \dfrac{1}{u^2 + a^2}\, du = \dfrac{1}{a}\arctan\dfrac{u}{a} + C

with

u = x + 2

and

a = 2

Pitfall: Forgetting the factor in front of arctan

Finishing with

\arctan(x + 2) + C

uses the formula for

a = 1

: its derivative is

\frac{1}{(x + 2)^2 + 1}

, not

\frac{1}{(x + 2)^2 + 4}

. With

a = 2

the answer is

\frac{1}{2}\arctan\frac{x + 2}{2} + C

Completing the square is the move for a quadratic that does not factor. The discriminant of

x^2 + 4x + 8

16 - 32 = -16 < 0

, so there are no real roots to split the fraction with. Writing

x^2 + 4x + 8 = (x + 2)^2 + 4

puts the variable part into

u^2

and leaves the constant

4 = 2^2

a^2

.

Rewriting constants exposes patterns:

Recognizing Standard Forms

\int \frac{1}{\sqrt{9 - x^2}}\, dx = \int \frac{1}{\sqrt{3^2 - x^2}}\, dx = \arcsin\frac{x}{3} + C

Setting Up Definite Integrals

For definite integrals, correct setup is essential.

Identify the variable: What quantity varies, and over what range?

Express the integrand: Write the quantity being accumulated in terms of the integration variable.

Determine bounds: Where does accumulation begin and end?

Example

Find the area under

y = x^2

from

x = 0

x = 3

\text{Area} = \int_0^3 x^2\, dx = \frac{x^3}{3}\Big|_0^3 = \frac{27}{3} - 0 = 9

Check reasonableness: the area should be positive and between

0 \cdot 3 = 0

and

9 \cdot 3 = 27

Handling Absolute Values

Absolute values require splitting the integral where the argument changes sign.

Handling Absolute Values

\int_{-2}^{3} |x|\, dx

Since

|x| = -x

for

x < 0

and

|x| = x

for

x \geq 0

Handling Absolute Values

= \int_{-2}^{0} (-x)\, dx + \int_0^3 x\, dx

Handling Absolute Values

= \left[-\frac{x^2}{2}\right]_{-2}^{0} + \left[\frac{x^2}{2}\right]_0^3 = (0 - (-2)) + \left(\frac{9}{2} - 0\right) = 2 + \frac{9}{2} = \frac{13}{2}

Note:

\left|\int f\right| \neq \int |f|

in general.

Splitting at the sign change

The section’s ∫ from −2 to 3 of |x| dx. Left of 0 the integrand is −x and the region is a triangle with base 2 and height 2, area 2 (amber). Right of 0 it is x and the triangle has base 3 and height 3, area 9/2 (blue). The integral is the sum of the two pieces, 13/2; a single antiderivative over the whole interval would miss the change of formula at 0.

Integrating x over the whole interval would give 5/2: the left triangle would count as negative.

Piecewise Functions

Split the integral at boundaries between pieces and apply additivity:

Piecewise Functions

\int_a^c f(x)\, dx = \int_a^b f(x)\, dx + \int_b^c f(x)\, dx

Example

For

f(x) = \begin{cases} x^2 & x < 1 \\ 2x & x \geq 1 \end{cases}

, evaluate

\int_0^2 f(x)\, dx

= \int_0^1 x^2\, dx + \int_1^2 2x\, dx = \frac{x^3}{3}\Big|_0^1 + x^2\Big|_1^2

= \frac{1}{3} + (4 - 1) = \frac{1}{3} + 3 = \frac{10}{3}

Pitfall: Using one formula on the whole interval

Integrating

x^2

over all of

[0, 2]

gives

\frac{8}{3}

, and using

2x

throughout gives

4

. Neither is the integral of

f

, which is

\frac{1}{3} + 3 = \frac{10}{3}

The boundaries between the pieces are the only places where the integral has to be cut, and the number of pieces sets the number of integrals to add. The value of

f

at the single point

x = 1

does not affect either integral, so it makes no difference which piece the boundary point belongs to. A function with three or more pieces is handled the same way, with one integral per piece.

Symmetry Shortcuts

Symmetric integrands over symmetric intervals simplify dramatically.

Even functions satisfy

f(-x) = f(x)

Even Function Symmetry

\int_{-a}^{a} f(x)\, dx = 2 \int_0^a f(x)\, dx \quad \text{when } f(-x) = f(x)

Learn more about this formula: Even Function Symmetry →

Odd functions satisfy

f(-x) = -f(x)

Odd Function Symmetry

\int_{-a}^{a} f(x)\, dx = 0 \quad \text{when } f(-x) = -f(x)

Learn more about this formula: Odd Function Symmetry →

Browse all integral formulas →

Example

\int_{-3}^{3} x^3\, dx = 0 \quad \text{(odd function)}

\int_{-2}^{2} x^4\, dx = 2\int_0^2 x^4\, dx = 2 \cdot \frac{32}{5} = \frac{64}{5} \quad \text{(even function)}

Pitfall: Missing the even part of the integrand

Seeing

x^3

\int_{-2}^{2} (x^3 + 1)\, dx

and answering

0

ignores the even term

1

. Only the

x^3

part cancels, so the integral is

0 + 2 \cdot 2 = 4

Both shortcuts rest on comparing

f(-x)

with

f(x)

. In the first example

(-x)^3 = -x^3

, so the region on

[-3, 0]

is the region on

[0, 3]

reflected below the axis. In the second

(-x)^4 = x^4

, and the two halves are equal. Replacing

x

-x

takes one line and settles which shortcut, if any, applies.

Odd symmetry cancels

The section’s ∫ from −3 to 3 of x³ dx. x³ is odd, so the region on [−3, 0] is the region on [0, 3] turned upside down: the same area, 81/4, but below the axis. The two signed areas cancel and the integral is 0 without any antiderivative. For an even function the two halves would match with the same sign, giving twice the right-hand half.

The shortcut needs a symmetric interval: over [−3, 2] the halves no longer match.

Structure in integrand	Recognize when…	Strategy
Absolute value	\|g(x)\| appears in the integrand	split the interval where g(x) = 0; use the sign of g on each piece to drop the bars
Piecewise function	different formulas govern different subintervals	split at the piece boundaries, integrate each subinterval, add via additivity
Even function on [−a, a]	f(−x) = f(x), symmetric interval	∫_−a^a f(x) dx = 2 ∫₀^a f(x) dx — half the work
Odd function on [−a, a]	f(−x) = −f(x), symmetric interval	∫_−a^a f(x) dx = 0 — answer is immediate

Common Pitfalls

Forgetting the constant of integration: Indefinite integrals always include

+C

.

Dropping absolute values: The antiderivative of

1/x

\ln|x| + C

, not

\ln x + C

.

Sign errors in substitution: When

u = -x

, then

du = -dx

, not

dx

.

Forgetting to convert limits: In definite integrals with substitution, either convert the limits to

u

-values or substitute back to

x

before evaluating.

Missing discontinuities: An integral like

\int_{-1}^{1} \dfrac{1}{x^2}\, dx

is improper—the integrand is unbounded at

x = 0

.

Integrating the top and bottom separately

A fraction is integrated by finding an antiderivative of the numerator and one of the denominator and dividing the two, as if integration had a quotient rule that works that way.

Wrong working:

\int \dfrac{x^3 + 1}{x}\, dx = \dfrac{\int (x^3 + 1)\, dx}{\int x\, dx} = \dfrac{x^4/4 + x}{x^2/2} = \dfrac{x^2}{2} + \dfrac{2}{x}

.

Right working: split the fraction first.

\dfrac{x^3 + 1}{x} = x^2 + \dfrac{1}{x}

, so

\int \dfrac{x^3 + 1}{x}\, dx = \dfrac{x^3}{3} + \ln|x| + C

.

Why it happens: an integral does split over a sum, term by term, and the same splitting looks just as natural over a quotient. There is no such rule for quotients; even for derivatives, the quotient rule is not the quotient of the derivatives.

How to check: differentiate the answer.

\left(\dfrac{x^2}{2} + \dfrac{2}{x}\right)' = x - \dfrac{2}{x^2}

, which is

-1

x = 1

, while the integrand there is

\dfrac{1 + 1}{1} = 2

\left(\dfrac{x^3}{3} + \ln|x|\right)' = x^2 + \dfrac{1}{x}

, which is

2

x = 1

.

Forgetting the factor in front of arctan

Completing the square brings the integral to the form

\int \dfrac{1}{u^2 + a^2}\, du

, and it is finished as

\arctan u + C

, which is the formula for

a = 1

only.

Wrong working:

\int \dfrac{1}{x^2 + 4x + 8}\, dx = \int \dfrac{1}{(x + 2)^2 + 4}\, dx = \arctan(x + 2) + C

.

Right working: with

u = x + 2

and

a = 2

\int \dfrac{1}{(x + 2)^2 + 4}\, dx = \dfrac{1}{2}\arctan\dfrac{x + 2}{2} + C

.

Why it happens:

\int \dfrac{1}{1 + u^2}\, du = \arctan u + C

is the version learned first, and for

a = 1

both the factor

\dfrac{1}{a}

and the division inside the arctan disappear.

How to check: differentiate.

\left(\arctan(x + 2)\right)' = \dfrac{1}{(x + 2)^2 + 1} = \dfrac{1}{x^2 + 4x + 5}

, the wrong constant.

\left(\dfrac{1}{2}\arctan\dfrac{x + 2}{2}\right)' = \dfrac{1}{2} \cdot \dfrac{1/2}{1 + \frac{(x + 2)^2}{4}} = \dfrac{1}{(x + 2)^2 + 4}

, the integrand.

Using one formula on the whole interval

A piecewise function is integrated with one of its formulas over the whole interval, usually the first one or the easier one, and the boundary between the pieces is ignored.

Wrong working: for

f(x) = x^2

when

x < 1

and

f(x) = 2x

when

x \geq 1

\int_0^2 f(x)\, dx = \int_0^2 x^2\, dx = \dfrac{8}{3}

. Using the other formula throughout gives

\int_0^2 2x\, dx = 4

, just as wrong.

Right working: split at

x = 1

\int_0^1 x^2\, dx + \int_1^2 2x\, dx = \dfrac{1}{3} + 3 = \dfrac{10}{3}

.

Why it happens: the integral is written with a single symbol

f

, so

\int_0^2 f(x)\, dx

looks like one integral. The cases are read once at the start and then forgotten.

How to check: list the subintervals used and match each one to the case of

f

that holds there. Integrating

x^2

[1, 2]

uses the wrong formula: at

x = 1.5

, for instance,

f(1.5) = 3

, not

1.5^2 = 2.25

.

Missing the even part of the integrand

An integrand on a symmetric interval contains an odd term such as

x^3

, and the whole integral is set to

0

, although another term is even.

Wrong working:

x^3

is odd, so

\int_{-2}^{2} (x^3 + 1)\, dx = 0

.

Right working: split by linearity.

\int_{-2}^{2} x^3\, dx + \int_{-2}^{2} 1\, dx = 0 + 2\int_0^2 1\, dx = 0 + 4 = 4

. Directly:

\left[\dfrac{x^4}{4} + x\right]_{-2}^{2} = (4 + 2) - (4 - 2) = 4

.

Why it happens: the odd rule gives the answer with no work, so it is tempting to use it as soon as an odd power appears, and the eye stops at the leading term.

How to check: test

f(-x) = -f(x)

for the whole integrand at one point. For

f(x) = x^3 + 1

f(-1) = 0

while

-f(1) = -2

, so

f

is not odd.

A positive integrand with a negative answer

The section’s ∫ from −1 to 1 of dx/x². The integrand is positive at every x ≠ 0, so any area it encloses must be positive. Plugging the endpoints into the antiderivative −1/x gives −1 − 1 = −2, which cannot be right. The fundamental theorem needs a continuous integrand, and 1/x² blows up at 0; treated as an improper integral, each half diverges.

Before evaluating, check that the integrand is continuous on the whole closed interval.

Pitfall	Why it's wrong	Correct form
Forgetting + C	an indefinite integral represents a family of antiderivatives, not just one	always finish indefinite results with + C
Dropping the absolute value in ln	∫ 1/x dx = ln x is invalid for x < 0	∫ 1/x dx = ln\|x\| + C
Sign error in u-substitution	if u = −x then du has its own sign	du = −dx, not dx
Forgetting to convert limits	x-limits are wrong once the integrand is in u	convert the limits to u-values, or substitute back to x before evaluating
Missing a discontinuity	integral looks proper but the integrand blows up on the interval	recognize as improper and split at the singularity (e.g. ∫₋₁¹ 1/x² dx at x = 0)

Checking Your Answer

Differentiate: The derivative of your antiderivative should return the integrand. This catches algebraic and sign errors.

Verify: For

\int \sec^2 x\, dx = \tan x + C

, check:

(\tan x)' = \sec^2 x

. Correct.

Estimate: For definite integrals, check that the answer is reasonable. A positive integrand on

[a,b]

with

a < b

should yield a positive result. The value should lie between

m(b-a)

and

M(b-a)

where

m

and

M

bound the integrand.

Numerical check: When possible, compare to a calculator or numerical approximation.

Summary: The Evaluation Strategy Ladder

The sections above lay out a natural order of attack for any integral: start simple, escalate only as needed, and verify at the end. The table below collects this six-step ladder with a pointer to the right section or sibling page for each step. Used as a checklist, it prevents the most common failure mode of integration — reaching for heavy techniques before trying the lighter moves that would have worked.

Step	What to do	Why this step	Where to look
1	Simplify the integrand — expand, split fractions, separate terms	algebraic cleanup often reveals matches to known formulas	section 1 above
2	Apply direct antidifferentiation — match the integrand to a standard form	fastest path when it works	special integrals
3	Look for disguised standard forms — complete the square, rewrite constants	catches near-misses before reaching for heavier tools	section 2 above
4	Check for special structure — absolute value, piecewise, symmetry	dedicated shortcuts avoid heavy machinery (and the odd-function rule answers in one step)	sections 4–6 above
5	Reach for a technique — substitution, parts, partial fractions, trig sub	when none of the above moves close the gap	techniques
6	Verify — differentiate the antiderivative; check sign and reasonableness for definite results	catches algebra and sign errors before they propagate	sections 7 & 8 above

Evaluating Integrals FAQ

How do you integrate a fraction like (x³ + 1)/x?

Divide term by term before integrating. (x³ + 1)/x becomes x² + 1/x, and each piece is a standard form, giving x³/3 + ln|x| + C. There is no quotient rule for integration, so a fraction usually has to be rewritten as a sum, expanded, or split before any formula applies.Read more →

When do you complete the square to evaluate an integral?

When the integrand holds an irreducible quadratic that almost matches an arctan or arcsin form. Rewriting x² + 4x + 8 as (x + 2)² + 4 turns ∫1/(x² + 4x + 8) dx into the arctan pattern with u = x + 2 and a = 2. The same move rewrites 9 − x² as 3² − x² to expose arcsin.Read more →

Is the absolute value of an integral the same as the integral of the absolute value?

No — |∫f| ≤ ∫|f|, and they are usually unequal. Integrating f lets positive and negative regions cancel before you take the absolute value; integrating |f| flips the negative parts up first, so nothing cancels. For f(x) = x on [−1, 1], the first is 0 and the second is 1.Read more →

Why is the integral of an odd function over a symmetric interval zero?

Because the two halves cancel exactly. An odd function satisfies f(−x) = −f(x), so whatever signed area accumulates on [0, a] is matched by the negative of it on [−a, 0], and the sum is 0 for every a. Even functions do the opposite: the halves match in sign, so the integral is twice the half.Read more →