Basic integration rules — power, trig, exponential, log: Question bank

The figure above shows the hole at $n = - 1$ : the power-rule denominator $n + 1$ passes through zero, so the formula blows up exactly where $ln ∣ x ∣$ has to take over.

True or false — justify

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C

holds for every real number

n

False — it fails at

n = - 1

, where

n + 1 = 0

and you'd divide by zero; that single hole is filled by

\int x^{- 1} d x = ln ∣ x ∣ + C

Two antiderivatives of the same function must be identical.

False — they can differ by any constant, since differentiation flattens constants to

0

; e.g.

x^{2}

and

x^{2} + 7

both have derivative

2 x

\int sin x d x = cos x + C

False — it is

- cos x + C

; check by differentiating:

\frac{d}{d x} (- cos x) = sin x

, whereas

\frac{d}{d x} cos x = - sin x

(wrong sign).

\int e^{x} d x = e^{x} + C

because

e^{x}

is unchanged by integration.

True —

e^{x}

is the fixed point of differentiation (

\frac{d}{d x} e^{x} = e^{x}

), so read backwards it is also unchanged by integration.

\int a^{x} d x = a^{x} + C

for any base

a > 0

False — since

\frac{d}{d x} a^{x} = a^{x} ln a

, that answer is

ln a

times too big; the correct integral is

\frac{a ^{x}}{l n a} + C

(only

a = e

leaves it unchanged, because

ln e = 1

You may never omit the

+ C

, not even in the middle of a calculation.

False (overstated) — in intermediate steps you may drop

+ C

for brevity, provided you restore it in the final answer; what you must never do is present a finished indefinite integral without it, since the answer is an entire family of functions, not one.

\int \frac{1}{x} d x = ln x + C

is the fully correct answer.

False (incomplete) —

ln x

is undefined for

x < 0

but

\frac{1}{x}

exists there; the correct answer

ln ∣ x ∣ + C

is valid for all

x \neq = 0

Linearity lets you write

\int f (x) g (x) d x = (\int f d x) (\int g d x)

False — linearity comes from the derivative sum rule

(f + g)^{'} = f^{'} + g^{'}

, which reverses cleanly; but the derivative product rule

(f g)^{'} = f^{'} g + f g^{'}

has an extra cross-term, so it does not reverse into a product of integrals — products need Integration by parts or Integration by substitution.

\int 5 d x = 5 x + C

True — treat

5

5 x^{0}

; the power rule gives

5 \cdot \frac{x ^{1}}{1} = 5 x

, or just note that

\frac{d}{d x} (5 x) = 5

Spot the error

\int x^{- 3} d x = \frac{x ^{- 2}}{- 3} + C

Wrong denominator — you divide by the new power

n + 1 = - 2

, not the old one; the answer is

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

\int sec^{2} x d x = sec x tan x + C

That's confusing an integral with a derivative;

\frac{d}{d x} tan x = sec^{2} x

, so

\int sec^{2} x d x = tan x + C

\int 2^{x} d x = \frac{2 ^{x + 1}}{x + 1} + C

The power rule was misapplied to an exponential — here

x

is the exponent, not the base, so the power rule doesn't apply at all; use

\int a^{x} d x = \frac{a ^{x}}{l n a} + C = \frac{2 ^{x}}{l n 2} + C

\int (3 x^{2}) (x) d x = (\int 3 x^{2} d x) (\int x d x) = x^{3} \cdot \frac{x ^{2}}{2}

Invalid — you cannot integrate a product term-by-term as a product; first multiply out to

3 x^{3}

, then integrate to get

\frac{3 x ^{4}}{4} + C

\int cos x d x = - sin x + C

Sign error — the minus belongs to the sine integral, not the cosine one;

\int cos x d x = sin x + C

since

\frac{d}{d x} sin x = cos x

\int \frac{4}{x} d x = 4 \cdot \frac{x ^{0}}{0} + C

Illegal division by zero — this is exactly the

n = - 1

case where the power rule breaks; pull out the

4

and use the log rule:

4 ln ∣ x ∣ + C

"For

x \geq 0

\int x d x = \frac{2}{3} (x)^{3} + C

, and also

= \frac{x ^{1/2 + 1}}{1/2} + C

The first is correct — writing it as an exponent,

(x)^{3} = x^{3/2}

, so it equals

\frac{2}{3} x^{3/2} + C

on the domain

x \geq 0

where

x

lives; but the second divides by the old power

\frac{1}{2}

instead of the new one

\frac{3}{2}

, so it's wrong (the correct denominator is

\frac{3}{2}

, giving

\frac{x ^{3/2}}{3/2} = \frac{2}{3} x^{3/2}

Why questions

Why must a

+ C

appear on every indefinite integral, but not on a definite integral?

Indefinite integrals name a whole family (the lost constant is unknown); in a definite integral you evaluate an antiderivative

F

at the endpoints, and the constant cancels in

F (b) - F (a)

, so it never appears — see Definite integrals & FTC.

Why does the power rule "raise then divide" rather than "lower then multiply"?

Because differentiation lowers the power and multiplies by it; integration reverses that, so we raise the power and divide by the new power to cancel the factor differentiation would produce.

Why is

e^{x}

singled out among all exponentials

a^{x}

Because

ln e = 1

, the correction factor

\frac{1}{l n a}

becomes

1

, so

e^{x}

integrates to itself; every other base carries a

\frac{1}{l n a}

— this is what makes

e

the "natural" base, see Natural log and exponential functions.

Why does a minus sign appear in

\int sin x d x

but not in

\int cos x d x

Reading the derivatives backwards:

\frac{d}{d x} cos x = - sin x

, so to get

sin x

you need

- cos x

; whereas

\frac{d}{d x} sin x = cos x

needs no adjustment.

Why is

ln ∣ x ∣

(with absolute value) needed instead of just

ln x

Because

\frac{1}{x}

is defined for negative

x

too, but

ln x

is not; the absolute value extends the antiderivative to all

x \neq = 0

while keeping

\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x}

(see the two-branch picture at the top).

Why is there no product or quotient rule for integration?

Because the product rule for derivatives,

(f g)^{'} = f^{'} g + f g^{'}

, has a sum of two terms; reversing it can't be undone in one clean formula — it turns into Integration by parts, and reversing chain-rule products turns into Integration by substitution; neither is as tidy as linearity.

Why do we always verify an integral by differentiating rather than re-integrating?

Differentiation is deterministic and unique, so it's a clean check; if

\frac{d}{d x} (answer)

equals the integrand, the answer is correct up to the unavoidable

+ C

Edge cases

What is

\int x^{0} d x

, and does the power rule apply?

Yes —

x^{0} = 1

, and

n = 0 \neq = - 1

, so

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

; the rule is safe because we're not dividing by zero.

What happens to the power rule exactly at

n = - 1

, and what replaces it?

The denominator

n + 1

becomes

0

, so the formula is undefined; nature patches the hole with an entirely different function,

ln ∣ x ∣ + C

(this is the blow-up shown in the second figure).

\int 1^{x} d x = \frac{1 ^{x}}{l n 1} + C

valid?

No —

ln 1 = 0

, so this divides by zero; but

1^{x} = 1

for all

x

, so just integrate the constant:

\int 1 d x = x + C

x = 0

, is

\int \frac{1}{x} d x = ln ∣ x ∣ + C

meaningful?

No —

\frac{1}{x}

isn't defined at

x = 0

, so the antiderivative is only valid on intervals not containing

0

; you cannot integrate "across" the origin here.

Can two antiderivatives on disconnected domains differ by different constants on each piece?

Yes — since

\frac{1}{x}

lives on two separate intervals (

x < 0

and

x > 0

) with a gap at

0

, the true antiderivative is

ln ∣ x ∣ + C_{1}

for

x > 0

and

ln ∣ x ∣ + C_{2}

for

x < 0

, with independent constants; writing a single "

+ C

" quietly assumes the same shift on both branches, which is the mild abuse (see the two-branch picture at the top).

What is

\int 0 d x

C

— the zero function is the derivative of any constant, so its antiderivative is just an arbitrary constant with no

x

term.

Connections

Basic integration rules — power, trig, exponential, log (the rules these traps stress-test)
Derivatives — basic rules (every trap resolves by differentiating backwards)
Integration by substitution (what replaces a "product rule" for composites)
Integration by parts (what replaces a "product rule" for genuine products)
Definite integrals & FTC (where the $+ C$ finally disappears)
Natural log and exponential functions (source of the $ln ∣ x ∣$ and $e^{x}$ subtleties)
Trigonometric integrals (extends the sign-sensitive trig cases)