Indefinite Integrals and Notation

Understanding integral notation and basic integration rules

∫ Indefinite Integrals and Notation

The Integral Symbol

The process of finding antiderivatives is called integration, and we use the integral symbol:

f(x)dx\int f(x)\,dx

This is read as "the integral of f(x)f(x) with respect to xx."


Notation Breakdown

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

Components:

  • \int - integral symbol (looks like an elongated S for "sum")
  • f(x)f(x) - the integrand (function being integrated)
  • dxdx - tells us we're integrating with respect to xx
  • F(x)+CF(x) + C - the antiderivative (general solution)

Indefinite vs. Definite Integrals

Indefinite Integral

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

  • Represents the family of all antiderivatives
  • Includes the constant +C+C
  • Result is a function

Definite Integral (coming later!)

abf(x)dx=F(b)F(a)\int_a^b f(x)\,dx = F(b) - F(a)

  • Has limits of integration (aa to bb)
  • Represents a specific number (area)
  • No +C+C needed

For now, we focus on indefinite integrals!


Basic Integration Formulas

Power Rule

xndx=xn+1n+1+C(n1)\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)

Examples:

  • x5dx=x66+C\int x^5\,dx = \frac{x^6}{6} + C
  • xdx=x22+C\int x\,dx = \frac{x^2}{2} + C
  • 1dx=x+C\int 1\,dx = x + C

Special Case: n=1n = -1

1xdx=lnx+C\int \frac{1}{x}\,dx = \ln|x| + C

Why absolute value?

  • lnx\ln x only defined for x>0x > 0
  • lnx\ln|x| works for all x0x \neq 0

Constant Rule

kdx=kx+C\int k\,dx = kx + C

where kk is any constant.

Example: 5dx=5x+C\int 5\,dx = 5x + C


Exponential Function

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

The exponential function is its own integral!


Trigonometric Functions

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

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

sec2xdx=tanx+C\int \sec^2 x\,dx = \tan x + C

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

csc2xdx=cotx+C\int \csc^2 x\,dx = -\cot x + C

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

Remember: Sine → negative cosine, Cosine → positive sine


Properties of Integrals

Constant Multiple Rule

kf(x)dx=kf(x)dx\int k \cdot f(x)\,dx = k \int f(x)\,dx

You can "pull out" constants!

Example: 3x2dx=3x2dx=3x33+C=x3+C\int 3x^2\,dx = 3\int x^2\,dx = 3 \cdot \frac{x^3}{3} + C = x^3 + C


Sum/Difference Rule

[f(x)±g(x)]dx=f(x)dx±g(x)dx\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx

Integrate term by term!

Example: (x2+3x)dx=x2dx+3xdx\int (x^2 + 3x)\,dx = \int x^2\,dx + \int 3x\,dx =x33+3x22+C= \frac{x^3}{3} + \frac{3x^2}{2} + C


Important: The dxdx

The "dxdx" is not optional - it tells us the variable of integration!

Example: t2dt\int t^2\,dt vs x2dx\int x^2\,dx

Same form, different variables!

Why it matters:

  • Clarifies which variable we're integrating
  • Essential for u-substitution (later topic)
  • Part of the mathematical notation

Working with Integrals

Example 1: Polynomial

Evaluate (4x32x2+5)dx\int (4x^3 - 2x^2 + 5)\,dx

Solution:

Apply sum rule and constant multiple rule:

=4x3dx2x2dx+5dx= \int 4x^3\,dx - \int 2x^2\,dx + \int 5\,dx

=4x3dx2x2dx+5dx= 4\int x^3\,dx - 2\int x^2\,dx + \int 5\,dx

=4x442x33+5x+C= 4 \cdot \frac{x^4}{4} - 2 \cdot \frac{x^3}{3} + 5x + C

=x42x33+5x+C= x^4 - \frac{2x^3}{3} + 5x + C


Example 2: Rewriting Before Integrating

Evaluate 3x4dx\int \frac{3}{x^4}\,dx

Step 1: Rewrite using exponents

3x4dx=3x4dx\int \frac{3}{x^4}\,dx = \int 3x^{-4}\,dx

Step 2: Apply power rule

=3x4+14+1+C=3x33+C= 3 \cdot \frac{x^{-4+1}}{-4+1} + C = 3 \cdot \frac{x^{-3}}{-3} + C

=33x3+C=1x3+C= -\frac{3}{3x^3} + C = -\frac{1}{x^3} + C


Example 3: Expanding First

Evaluate (x+1)2dx\int (x+1)^2\,dx

WRONG approach: Try to integrate directly ❌

RIGHT approach: Expand first!

(x+1)2dx=(x2+2x+1)dx\int (x+1)^2\,dx = \int (x^2 + 2x + 1)\,dx

=x33+2x22+x+C= \frac{x^3}{3} + \frac{2x^2}{2} + x + C

=x33+x2+x+C= \frac{x^3}{3} + x^2 + x + C


Example 4: Trigonometric

Evaluate (3sinx2cosx)dx\int (3\sin x - 2\cos x)\,dx

Solution:

=3sinxdx2cosxdx= 3\int \sin x\,dx - 2\int \cos x\,dx

=3(cosx)2(sinx)+C= 3(-\cos x) - 2(\sin x) + C

=3cosx2sinx+C= -3\cos x - 2\sin x + C


Fractional and Negative Exponents

Square Roots

xdx=x1/2dx=x3/23/2+C=2x3/23+C\int \sqrt{x}\,dx = \int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} + C = \frac{2x^{3/2}}{3} + C

Remember: x=x1/2\sqrt{x} = x^{1/2}


Cube Roots

x23dx=x2/3dx=x5/35/3+C=3x5/35+C\int \sqrt[3]{x^2}\,dx = \int x^{2/3}\,dx = \frac{x^{5/3}}{5/3} + C = \frac{3x^{5/3}}{5} + C


Reciprocals

1x2dx=x2dx=x11+C=1x+C\int \frac{1}{x^2}\,dx = \int x^{-2}\,dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C


Combining Multiple Techniques

Example: Mixed Terms

Evaluate (x2+2x2+x)dx\int \left(x^2 + \frac{2}{x^2} + \sqrt{x}\right)dx

Step 1: Rewrite everything as powers

=(x2+2x2+x1/2)dx= \int \left(x^2 + 2x^{-2} + x^{1/2}\right)dx

Step 2: Integrate term by term

=x33+2x11+x3/23/2+C= \frac{x^3}{3} + 2 \cdot \frac{x^{-1}}{-1} + \frac{x^{3/2}}{3/2} + C

=x332x+2x3/23+C= \frac{x^3}{3} - \frac{2}{x} + \frac{2x^{3/2}}{3} + C


The "+C+C" Convention

When combining multiple constants, we can consolidate:

(x2+1)dx=x33+C1+x+C2\int (x^2 + 1)\,dx = \frac{x^3}{3} + C_1 + x + C_2

We typically write this as: =x33+x+C= \frac{x^3}{3} + x + C

where C=C1+C2C = C_1 + C_2 (arbitrary constant).

One +C is enough at the end!


⚠️ Common Mistakes

Mistake 1: Forgetting +C

WRONG: x2dx=x33\int x^2\,dx = \frac{x^3}{3}

RIGHT: x2dx=x33+C\int x^2\,dx = \frac{x^3}{3} + C

Always include the constant of integration!


Mistake 2: Wrong Power Rule

WRONG: x3dx=x33+C\int x^3\,dx = \frac{x^3}{3} + C (forgot to add 1 to exponent)

RIGHT: x3dx=x44+C\int x^3\,dx = \frac{x^4}{4} + C


Mistake 3: Forgetting dxdx

WRONG: x2\int x^2

RIGHT: x2dx\int x^2\,dx

The dxdx is part of the notation!


Mistake 4: Can't Integrate Products/Quotients Directly

WRONG: (xx2)dx=(xdx)(x2dx)\int (x \cdot x^2)\,dx = \left(\int x\,dx\right) \cdot \left(\int x^2\,dx\right)

RIGHT: Simplify first! x3dx=x44+C\int x^3\,dx = \frac{x^4}{4} + C

There's no "product rule" for integrals!


Mistake 5: Sine/Cosine Signs

WRONG: sinxdx=cosx+C\int \sin x\,dx = \cos x + C

RIGHT: sinxdx=cosx+C\int \sin x\,dx = -\cos x + C (negative!)


When Basic Rules Don't Work

Some integrals need advanced techniques:

Can't do easily:

  • ex2dx\int e^{x^2}\,dx (needs special functions)
  • sin(x2)dx\int \sin(x^2)\,dx (needs special functions)
  • 1x2+1dx\int \frac{1}{x^2+1}\,dx (this is arctanx+C\arctan x + C, learned later)

Can do with techniques:

  • xcos(x2)dx\int x\cos(x^2)\,dx (u-substitution)
  • xexdx\int x e^x\,dx (integration by parts)

We'll learn these methods in future lessons!


Quick Reference Table

| Integrand | Integral | |-----------|----------| | xnx^n (n1n \neq -1) | xn+1n+1+C\frac{x^{n+1}}{n+1} + C | | 1x\frac{1}{x} | lnx+C\ln|x| + C | | exe^x | ex+Ce^x + C | | sinx\sin x | cosx+C-\cos x + C | | cosx\cos x | sinx+C\sin x + C | | sec2x\sec^2 x | tanx+C\tan x + C | | secxtanx\sec x \tan x | secx+C\sec x + C |


📝 Practice Strategy

  1. Rewrite the integrand if needed (expand, use exponents)
  2. Apply linearity (split sums, pull out constants)
  3. Use basic formulas (power rule, trig, exponential)
  4. Always include +C
  5. Check your answer by differentiating
  6. Include dxdx in your notation
  7. Simplify before integrating when possible

📚 Practice Problems

1Problem 1easy

Question:

Find the following indefinite integrals:

a) 6x5dx\int 6x^5 \, dx b) (3x24x+5)dx\int (3x^2 - 4x + 5) \, dx c) 1x3dx\int \frac{1}{x^3} \, dx

💡 Show Solution

Solution:

Part (a): Power rule for integration: xndx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C

6x5dx=6x66+C=x6+C\int 6x^5 \, dx = 6 \cdot \frac{x^6}{6} + C = x^6 + C

Part (b): Integrate term by term:

(3x24x+5)dx=3x334x22+5x+C\int (3x^2 - 4x + 5) \, dx = 3 \cdot \frac{x^3}{3} - 4 \cdot \frac{x^2}{2} + 5x + C

=x32x2+5x+C= x^3 - 2x^2 + 5x + C

Part (c): Rewrite with negative exponent: x3dx\int x^{-3} \, dx

=x22+C=12x2+C= \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C

2Problem 2easy

Question:

Evaluate (5x43x2+7x2)dx\int (5x^4 - 3x^2 + 7x - 2)\,dx.

💡 Show Solution

Step 1: Apply sum/difference and constant multiple rules

(5x43x2+7x2)dx\int (5x^4 - 3x^2 + 7x - 2)\,dx

=5x4dx3x2dx+7xdx2dx= 5\int x^4\,dx - 3\int x^2\,dx + 7\int x\,dx - \int 2\,dx


Step 2: Apply power rule to each term

=5x553x33+7x222x+C= 5 \cdot \frac{x^5}{5} - 3 \cdot \frac{x^3}{3} + 7 \cdot \frac{x^2}{2} - 2x + C


Step 3: Simplify

=x5x3+7x222x+C= x^5 - x^3 + \frac{7x^2}{2} - 2x + C


Check by differentiating:

ddx[x5x3+7x222x+C]\frac{d}{dx}\left[x^5 - x^3 + \frac{7x^2}{2} - 2x + C\right]

=5x43x2+7x2= 5x^4 - 3x^2 + 7x - 2

Answer: x5x3+7x222x+Cx^5 - x^3 + \frac{7x^2}{2} - 2x + C

3Problem 3easy

Question:

Find the following indefinite integrals:

a) 6x5dx\int 6x^5 \, dx b) (3x24x+5)dx\int (3x^2 - 4x + 5) \, dx c) 1x3dx\int \frac{1}{x^3} \, dx

💡 Show Solution

Solution:

Part (a): Power rule for integration: xndx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C

6x5dx=6x66+C=x6+C\int 6x^5 \, dx = 6 \cdot \frac{x^6}{6} + C = x^6 + C

Part (b): Integrate term by term:

(3x24x+5)dx=3x334x22+5x+C\int (3x^2 - 4x + 5) \, dx = 3 \cdot \frac{x^3}{3} - 4 \cdot \frac{x^2}{2} + 5x + C

=x32x2+5x+C= x^3 - 2x^2 + 5x + C

Part (c): Rewrite with negative exponent: x3dx\int x^{-3} \, dx

=x22+C=12x2+C= \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C

4Problem 4medium

Question:

Find (sinx+ex)dx\int (\sin x + e^x) \, dx.

💡 Show Solution

Solution:

Use basic integration formulas:

sinxdx=cosx+C1\int \sin x \, dx = -\cos x + C_1

exdx=ex+C2\int e^x \, dx = e^x + C_2

Combine:

(sinx+ex)dx=cosx+ex+C\int (\sin x + e^x) \, dx = -\cos x + e^x + C

where C=C1+C2C = C_1 + C_2

5Problem 5medium

Question:

Find (sinx+ex)dx\int (\sin x + e^x) \, dx.

💡 Show Solution

Solution:

Use basic integration formulas:

sinxdx=cosx+C1\int \sin x \, dx = -\cos x + C_1

exdx=ex+C2\int e^x \, dx = e^x + C_2

Combine:

(sinx+ex)dx=cosx+ex+C\int (\sin x + e^x) \, dx = -\cos x + e^x + C

where C=C1+C2C = C_1 + C_2

6Problem 6medium

Question:

Evaluate (4x3+2x3x)dx\int \left(\frac{4}{x^3} + 2\sqrt{x} - \frac{3}{x}\right)dx.

💡 Show Solution

Step 1: Rewrite using exponents

(4x3+2x3x)dx\int \left(\frac{4}{x^3} + 2\sqrt{x} - \frac{3}{x}\right)dx

=(4x3+2x1/23x1)dx= \int \left(4x^{-3} + 2x^{1/2} - 3x^{-1}\right)dx


Step 2: Integrate term by term

For 4x34x^{-3}: 4x3+13+1=4x22=42x2=2x24 \cdot \frac{x^{-3+1}}{-3+1} = 4 \cdot \frac{x^{-2}}{-2} = -\frac{4}{2x^2} = -\frac{2}{x^2}

For 2x1/22x^{1/2}: 2x1/2+11/2+1=2x3/23/2=22x3/23=4x3/232 \cdot \frac{x^{1/2+1}}{1/2+1} = 2 \cdot \frac{x^{3/2}}{3/2} = 2 \cdot \frac{2x^{3/2}}{3} = \frac{4x^{3/2}}{3}

For 3x1=3x-3x^{-1} = -\frac{3}{x}: 3lnx-3\ln|x|


Step 3: Combine

=2x2+4x3/233lnx+C= -\frac{2}{x^2} + \frac{4x^{3/2}}{3} - 3\ln|x| + C

Or equivalently: =2x2+4xx33lnx+C= -2x^{-2} + \frac{4x\sqrt{x}}{3} - 3\ln|x| + C

Answer: 2x2+4x3/233lnx+C-\frac{2}{x^2} + \frac{4x^{3/2}}{3} - 3\ln|x| + C

7Problem 7medium

Question:

Evaluate (2sinx+3cosxex)dx\int (2\sin x + 3\cos x - e^x)\,dx.

💡 Show Solution

Step 1: Apply sum/difference rule

(2sinx+3cosxex)dx\int (2\sin x + 3\cos x - e^x)\,dx

=2sinxdx+3cosxdxexdx= 2\int \sin x\,dx + 3\int \cos x\,dx - \int e^x\,dx


Step 2: Apply basic integration formulas

For sinx\sin x: 2sinxdx=2(cosx)=2cosx2\int \sin x\,dx = 2(-\cos x) = -2\cos x

For cosx\cos x: 3cosxdx=3(sinx)=3sinx3\int \cos x\,dx = 3(\sin x) = 3\sin x

For exe^x: exdx=ex-\int e^x\,dx = -e^x


Step 3: Combine and add +C

=2cosx+3sinxex+C= -2\cos x + 3\sin x - e^x + C


Check by differentiating:

ddx[2cosx+3sinxex+C]\frac{d}{dx}[-2\cos x + 3\sin x - e^x + C]

=2(sinx)+3(cosx)ex= -2(-\sin x) + 3(\cos x) - e^x

=2sinx+3cosxex= 2\sin x + 3\cos x - e^x

Answer: 2cosx+3sinxex+C-2\cos x + 3\sin x - e^x + C