Day 16: Antiderivatives & Basic Integration

Basic Antiderivative Rules

An antiderivative of $f$ is any function $F$ with $F' = f$ . Before Riemann sums, the FTC, or u-substitution, you need the reverse of every derivative rule at your fingertips — this is the vocabulary the whole integration unit is built on.

Core Theorem

Every antiderivative differs by a constant, so

\displaystyle\int f(x)\,dx = F(x) + C

.
Power rule:

\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C

for

n \neq -1

;

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

.
Exponential:

\int e^x\,dx = e^x + C

;

\int e^{kx}\,dx = \dfrac{1}{k}e^{kx} + C

;

\int a^x\,dx = \dfrac{a^x}{\ln a} + C

.
Trig:

\int \sin x\,dx = -\cos x + C

;

\int \cos x\,dx = \sin x + C

;

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

;

\int \csc^2 x\,dx = -\cot x + C

;

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

;

\int \csc x\cot x\,dx = -\csc x + C

.
Integration is linear:

\int [c\,f \pm g] = c\int f \pm \int g

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Find

\displaystyle\int \left(6x^2 - 4x + 5\right) dx

.

How to think about it

Antidifferentiation is the power rule backwards: raise the exponent by one, divide by the new exponent. Do it term by term (integration is linear), treat the constant

5

as

5x^0 \to 5x

, and never drop the

+C

.

Key idea: Apply the power rule term by term. The antiderivative of a constant $5$ is $5x$ .

Full solution

1. Step 1: Antidifferentiate Each Term

\dfrac{6x^3}{3} - \dfrac{4x^2}{2} + 5x = 2x^3 - 2x^2 + 5x

.

2. Step 2: Add the Constant

\displaystyle\int (6x^2 - 4x + 5)\,dx = 2x^3 - 2x^2 + 5x + C

.

Worked Example 02Medium

Find

\displaystyle\int \left(\sqrt{x} + \dfrac{3}{x^2} - \dfrac{2}{x}\right) dx

.

How to think about it

You can't use the power rule until each term is written as

x^{\text{power}}

, so first turn roots and fractions into exponents. The one exception is

\dfrac{2}{x} = 2x^{-1}

: the power rule would divide by zero, so its antiderivative is

2\ln|x|

.

Key idea: Rewrite every term as a power of $x$ first: $x^{1/2}$ , $3x^{-2}$ , and $\frac{2}{x}$ (which is the $n = -1$ case).

Full solution

1. Step 1: Rewrite as Powers

\displaystyle\int \left(x^{1/2} + 3x^{-2} - \dfrac{2}{x}\right) dx

.

2. Step 2: Antidifferentiate

\dfrac{x^{3/2}}{3/2} + \dfrac{3x^{-1}}{-1} - 2\ln|x| = \dfrac{2}{3}x^{3/2} - \dfrac{3}{x} - 2\ln|x| + C

.

Step-by-Step SOP

1
Rewrite Before Integrating
Turn roots, fractions, and quotients into sums of powers of $x$ (with negative or fractional exponents).
2
Reverse the Matching Derivative Rule
Power rule up by one; $\frac{1}{x} \to \ln|x|$ ; each trig and exponential reversal from the table.
3
Always Write + C
An indefinite integral without the constant of integration loses a point.
4
Check by Differentiating
The derivative of your answer must return the original integrand.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Find

\displaystyle\int \left(4e^{2x} + \sec^2 x - \sin x\right) dx

.

Need a hint?

\int e^{kx}\,dx = \frac{1}{k}e^{kx}

. The other two are direct trig reversals.

Show solution

1. Step 1: Antidifferentiate Each Term

4 \cdot \dfrac{1}{2}e^{2x} + \tan x - (-\cos x) = 2e^{2x} + \tan x + \cos x

.

2. Step 2: Add C

2e^{2x} + \tan x + \cos x + C

.

Practice 02Medium

Find

\displaystyle\int \dfrac{x^3 - 2x + 1}{x}\,dx

.

Need a hint?

Divide each term of the numerator by

x

first — do not try to integrate a quotient directly.

Show solution

1. Step 1: Split the Fraction

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

.

2. Step 2: Antidifferentiate

\dfrac{x^3}{3} - 2x + \ln|x| + C

.

Common Pitfalls

⚠
Using the Power Rule on 1/x $\int x^{-1}\,dx$ is the one exception — it is $\ln|x| + C$ , not $\frac{x^0}{0}$ .
⚠
Forgetting the 1/k on e^{kx} $\int e^{3x}\,dx = \frac{1}{3}e^{3x} + C$ . Dropping the $\frac{1}{3}$ is the most common exponential slip.

Antiderivatives Giving Arctangent and Arcsine

Two derivative rules run in reverse to give inverse-trig antiderivatives: $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ and $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$ . Recognizing an integrand as one of these forms — sometimes after completing the square — is a tested AB skill.

Core Theorem

\displaystyle\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\!\left(\dfrac{x}{a}\right) + C

\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\!\left(\dfrac{x}{a}\right) + C

The special cases

a = 1

give

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

and

\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C

. If the denominator is a general quadratic, complete the square to force it into

x^2 + a^2

or

a^2 - x^2

form.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Find

\displaystyle\int \dfrac{dx}{x^2 + 9}

.

How to think about it

Key idea: This is the arctangent form with $a^2 = 9$ , so $a = 3$ .

Full solution

1. Step 1: Identify a

x^2 + 9 = x^2 + 3^2

, so

a = 3

.

2. Step 2: Apply the Formula

\dfrac{1}{3}\tan^{-1}\!\left(\dfrac{x}{3}\right) + C

.

Worked Example 02Easy

Find

\displaystyle\int \dfrac{dx}{\sqrt{4 - x^2}}

.

How to think about it

Key idea: Arcsine form with $a^2 = 4$ .

Full solution

1. Step 1: Identify a

4 - x^2 = 2^2 - x^2

, so

a = 2

.

2. Step 2: Apply the Formula

\sin^{-1}\!\left(\dfrac{x}{2}\right) + C

.

Step-by-Step SOP

1
Match the Shape
$\frac{1}{\text{quadratic}} \Rightarrow$ arctangent; $\frac{1}{\sqrt{\text{quadratic}}} \Rightarrow$ arcsine.
2
Read Off a
Write the constant as $a^2$ ; then $a$ appears both inside the inverse trig and (for arctangent) as a $\frac{1}{a}$ out front.
3
Complete the Square If Needed
A denominator like $x^2 + bx + c$ becomes $(x + b/2)^2 + (\text{constant})$ — then it is a shifted standard form.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Evaluate

\displaystyle\int_0^1 \dfrac{dx}{1 + x^2}

.

Need a hint?

The antiderivative is

\tan^{-1}x

; evaluate from 0 to 1.

Show solution

1. Step 1: Antiderivative

\displaystyle\int_0^1 \dfrac{dx}{1+x^2} = \Big[\tan^{-1}x\Big]_0^1

.

2. Step 2: Evaluate

\tan^{-1}(1) - \tan^{-1}(0) = \dfrac{\pi}{4} - 0 = \dfrac{\pi}{4}

.

Practice 02Hard

Find

\displaystyle\int \dfrac{dx}{x^2 - 6x + 13}

.

Need a hint?

The denominator does not factor over the reals. Complete the square:

x^2 - 6x + 13 = (x-3)^2 + 4

.

Show solution

1. Step 1: Complete the Square

x^2 - 6x + 13 = (x^2 - 6x + 9) + 4 = (x - 3)^2 + 4

.

2. Step 2: Recognize the Arctangent Form

\displaystyle\int \dfrac{dx}{(x-3)^2 + 2^2}

— the shifted arctangent form with

a = 2

.

3. Step 3: Apply the Formula

\dfrac{1}{2}\tan^{-1}\!\left(\dfrac{x - 3}{2}\right) + C

.

Common Pitfalls

⚠
Dropping the 1/a on the Arctangent $\int \frac{dx}{x^2 + a^2}$ has a $\frac{1}{a}$ factor. The arcsine form does not — do not add one there.
⚠
Trying to Factor a Sum of Squares $x^2 + 9$ does not factor over the reals. If a quadratic denominator has no real roots, it is an inverse-trig integral, not a partial-fractions one.

Indefinite Integrals and Initial-Value Problems

The $+C$ in an antiderivative is not a formality — it is a free parameter. One extra piece of information (a point on the curve, a starting position) pins it down and turns a family of curves into a single function. This is exactly how you recover position from acceleration.

Core Theorem

If

\dfrac{dy}{dx} = f(x)

and

y = y_0

when

x = x_0

, then

y = \displaystyle\int f(x)\,dx = F(x) + C

, and

C

is found by substituting the point:

y_0 = F(x_0) + C

.
For motion:

v(t) = \int a(t)\,dt

(with

C

from a known velocity), then

s(t) = \int v(t)\,dt

(with

C

from a known position).

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Find

f(x)

if

f'(x) = 6x^2 - 4

and

f(1) = 5

.

How to think about it

Key idea: Antidifferentiate to get $f(x) + C$ , then use $f(1) = 5$ to solve for $C$ .

Full solution

1. Step 1: Antidifferentiate

f(x) = 2x^3 - 4x + C

.

2. Step 2: Apply the Initial Condition

f(1) = 2 - 4 + C = 5 \implies C = 7

.

3. Step 3: Write the Solution

f(x) = 2x^3 - 4x + 7

.

Step-by-Step SOP

1
Antidifferentiate the Right-Hand Side
Get the general antiderivative $F(x) + C$ .
2
Substitute the Known Point
Replace $x$ and $y$ with the given values and solve the resulting equation for $C$ .
3
Repeat for Second-Order Problems
$f'' \to f'$ (find $C_1$ from $f'(x_0)$ ), then $f' \to f$ (find $C_2$ from $f(x_0)$ ).

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

A function satisfies

f''(x) = 12x

, with

f'(0) = 2

and

f(0) = -1

. Find

f(x)

.

Need a hint?

Antidifferentiate twice, finding a constant after each step.

Show solution

1. Step 1: First Antiderivative

f'(x) = 6x^2 + C_1

. Since

f'(0) = 2

,

C_1 = 2

, so

f'(x) = 6x^2 + 2

.

2. Step 2: Second Antiderivative

f(x) = 2x^3 + 2x + C_2

. Since

f(0) = -1

,

C_2 = -1

.

3. Step 3: Solution

f(x) = 2x^3 + 2x - 1

.

Practice 02Hard

A particle moving along a line has acceleration

a(t) = 6t - 2

(m/s²). At

t = 0

, its velocity is

v(0) = 3

m/s and its position is

s(0) = 1

m. Find

s(t)

.

Need a hint?

Integrate

a

for

v

(constant from

v(0)

), then integrate

v

for

s

(constant from

s(0)

).

Show solution

1. Step 1: Velocity from Acceleration

v(t) = \int (6t - 2)\,dt = 3t^2 - 2t + C_1

.

v(0) = 3 \implies C_1 = 3

, so

v(t) = 3t^2 - 2t + 3

.

2. Step 2: Position from Velocity

s(t) = \int (3t^2 - 2t + 3)\,dt = t^3 - t^2 + 3t + C_2

.

s(0) = 1 \implies C_2 = 1

.

3. Step 3: Solution

s(t) = t^3 - t^2 + 3t + 1

.

Common Pitfalls

⚠
Applying Both Conditions to the Same AntiderivativeIn a second-order problem, $f'(0)$ pins the constant on $f'$ , and $f(0)$ pins the constant on $f$ — one condition per integration step, in order.
⚠
Solving for C Before Antidifferentiating FullyFinish the whole antiderivative (all terms) before plugging in the point, or the value of $C$ will be wrong.

Worked Examples

1. Step 1: Antidifferentiate Each Term

2. Step 2: Add the Constant

1. Step 1: Rewrite as Powers

2. Step 2: Antidifferentiate

Rewrite Before Integrating

Reverse the Matching Derivative Rule

Always Write + C

Check by Differentiating

Practice — Your Turn

1. Step 1: Antidifferentiate Each Term

2. Step 2: Add C

1. Step 1: Split the Fraction

2. Step 2: Antidifferentiate

Worked Examples

1. Step 1: Identify a

2. Step 2: Apply the Formula

1. Step 1: Identify a

2. Step 2: Apply the Formula

Match the Shape

Read Off a

Complete the Square If Needed

Practice — Your Turn

1. Step 1: Antiderivative

2. Step 2: Evaluate

1. Step 1: Complete the Square

2. Step 2: Recognize the Arctangent Form

3. Step 3: Apply the Formula

Worked Examples

1. Step 1: Antidifferentiate

2. Step 2: Apply the Initial Condition

3. Step 3: Write the Solution

Antidifferentiate the Right-Hand Side

Substitute the Known Point

Repeat for Second-Order Problems

Practice — Your Turn

1. Step 1: First Antiderivative

2. Step 2: Second Antiderivative

3. Step 3: Solution

1. Step 1: Velocity from Acceleration

2. Step 2: Position from Velocity

3. Step 3: Solution

Gary Chang