Chapter 13 Week 4/5 In-Class: Integration Techniques

13.1 Purpose of This Workbook

This workbook is designed to help you learn integration by doing.

In lecture and the textbook, you are introduced to integration techniques as ideas:
how they work, why they exist, and how they connect back to derivatives.
This workbook is where those ideas become skills.

Rather than presenting long solution sets to copy, each section of this workbook is structured to support active problem solving:

  • short notes that highlight what to notice,
  • guided examples worked through together in class,
  • scaffolded practice that helps you choose a method deliberately,
  • independent problems that build confidence and fluency.

The emphasis throughout is not on memorizing formulas, but on recognizing structure and making informed strategic choices.

13.2 How to Use This Workbook

Each section follows the same rhythm:

  1. What to Notice
    Brief notes that highlight key patterns, warning signs, and strategy cues.

  2. Worked Example (Instructor-Led)
    A carefully chosen example you will work through together in class.
    These examples emphasize why a method works, not just how to execute it.

  3. Guided Practice
    One or two problems with prompts to help you identify structure before computing.

  4. Independent Practice
    A small set of problems to reinforce the technique and test your understanding.
    Some problems are designed to look tempting but require a different method.

  5. Strategy Check
    Short reflection questions that help you articulate how you chose a method.

You are encouraged to pause before computing and ask: > What structure do I see, and what method does it suggest?

That habit is more important than speed.

13.3 Learning Outcomes

By actively engaging with this workbook, you should be able to:

  • identify when an integral can be evaluated using basic antiderivative rules,
  • recognize nested structures and apply substitution appropriately,
  • identify product structures and use integration by parts strategically,
  • decompose rational functions using partial fractions and integrate term by term,
  • explain why a particular integration method is appropriate in a given situation,
  • distinguish between integrals that have elementary antiderivatives and those that do not,
  • check your results by differentiation and by interpreting units and sign,
  • develop confidence in rejecting inappropriate methods and course-correcting.

13.4 A Note on Productive Struggle

Some problems in this workbook are intentionally designed to fail if the wrong method is chosen.

This is not a trick.

Being able to recognize why a method does not apply—and to adjust your strategy—is a core calculus skill.
Mistakes, false starts, and revisions are part of learning to think mathematically.

If a problem feels uncomfortable at first, that is often a sign you are learning the right thing.

13.5 Expectations

To get the most out of this workbook:

  • write down your reasoning, not just final answers,
  • clearly indicate your choice of method before computing,
  • use differentiation to check your work whenever possible,
  • reflect briefly on strategy, even when the computation was straightforward.

Integration is not about finding the “right formula.”
It is about understanding structure, making good decisions, and refining your approach with practice.

This workbook is designed to help you build that skill set.

13.6 Section 1: Reverse Differentiation (Basic Antiderivative Rules)

Before learning new techniques, it is essential to be fluent with basic antiderivatives.
Most integration problems become difficult only because these foundations are shaky.

This section focuses on building the habit of recognition first, computation second.

13.6.1 What to Notice

  • Integration is reverse differentiation
  • Ask: “What function has this derivative?”
  • Constants factor out of integrals
  • Sums integrate term-by-term
  • Structure matters more than symbols
  • If an integrand matches a known derivative exactly, no advanced technique is needed

13.6.2 The Core Antiderivative Rules

These are entries in your antiderivative dictionary. You should recognize them immediately.

  • Power Rule (Reverse) \[ \int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) \]

  • Logarithmic Case \[ \int \frac{1}{x} dx = \ln|x| + C \]

  • Exponential \[ \int e^x dx = e^x + C \]

  • Constant Multiple \[ \int c\,f(x)\,dx = c\int f(x)\,dx \]

  • Linearity \[ \int [f(x)+g(x)]dx = \int f(x)dx + \int g(x)dx \]

13.6.3 Worked Example (Instructor-Led)

Evaluate: \[ \int 6x^5 \, dx \]

Pause & Predict

Before computing:

  • What derivative produces \(6x^5\)?
  • Do you need an advanced technique?

Recognize the Structure

From differentiation: \[ \frac{d}{dx}(x^6) = 6x^5 \]

So the antiderivative is immediate.

Integrate

\[ \int 6x^5 dx = x^6 + C \]

Why This Was Easy

Nothing was nested.
Nothing was multiplied in a way that required restructuring.
This integral matched a known derivative exactly.

This is the ideal situation—always check for it first.

13.6.4 Guided Practice (With Prompts)

Evaluate: \[ \int 4x^3 \, dx \]

Before computing:

  1. What power of \(x\) would differentiate to give \(4x^3\)?
  2. What constant must appear in the result?
  3. Does this require any technique beyond the power rule?

Now compute the integral.

13.6.5 Independent Practice

Evaluate each integral using basic rules only.

  1. \(\displaystyle \int x^7 dx\)

  2. \(\displaystyle \int 10x^2 dx\)

  3. \(\displaystyle \int x^{-3} dx\)

  4. \(\displaystyle \int \frac{1}{x} dx\)

  5. \(\displaystyle \int (3x^2 + 5x - 1) dx\)

(Use linearity and constant multiples where appropriate.)

13.6.6 Common Mistakes to Watch For

  • Forgetting the constant of integration
  • Misapplying the power rule when \(n = -1\)
  • Treating sums as products
  • Overcomplicating a problem that already matches a basic rule

If the integrand looks simple, trust that instinct.

13.6.7 Strategy Check

Answer briefly:

  • How can you tell when an integral can be handled immediately?
  • Why should basic rules always be checked before trying substitution or integration by parts?
  • Which integrals in the practice set required linearity?

Fluency with reverse differentiation is the foundation for everything that follows.

13.7 Section 2: Substitution (The Chain Rule in Reverse)

Many integrals are not difficult because they are complicated, but because they are nested. Substitution is the technique that allows us to temporarily simplify a nested structure by changing variables.

Conceptually, substitution is nothing more than the chain rule in reverse.

13.7.1 What to Notice

  • Substitution applies when a function is inside another function
  • Look for a pattern of the form: \[ f(g(x)) \cdot g'(x) \]
  • The derivative of the inner function must be:
    • present exactly ✔️
    • present up to a constant ✔️
    • missing ✖️
  • Substitution changes both the expression and the differential
  • If substitution does not simplify the integral, it is not the right tool

13.7.2 The Core Pattern

From differentiation, the chain rule says: \[ \frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x) \]

Reversing this idea gives the substitution pattern: \[ \int F'(g(x)) \cdot g'(x)\,dx = F(g(x)) + C \]

The key question is always:

Do I see a function and (almost) its derivative multiplied together?

13.7.3 Worked Example (Instructor-Led)

Evaluate: \[ \int (3x^2 + 1)^4 \cdot 6x \, dx \]

Pause & Predict

Before computing:

  • What is the inner function?
  • Do you see its derivative?

Identify the Structure

  • Inner function: \(3x^2 + 1\)
  • Derivative: \(6x\)

The integrand matches the chain rule pattern exactly.

Choose the Substitution

Let: \[ u = 3x^2 + 1 \]

Differentiate with respect to \(x\): \[ \frac{du}{dx} = 6x \quad \Rightarrow \quad du = 6x\,dx \]

Rewrite the Integral

Substitute:

  • \((3x^2 + 1) \rightarrow u\)
  • \(6x\,dx \rightarrow du\)

\[ \int u^4 \, du \]

Integrate

\[ \frac{u^5}{5} + C \]

Substitute back: \[ \frac{(3x^2 + 1)^5}{5} + C \]

Why This Worked

The integrand already was a derivative — it was just written in a nested form. Substitution removed the nesting so the basic power rule could be applied.

13.7.4 Guided Practice (With Prompts)

Evaluate: \[ \int \frac{4x}{x^2 + 7} \, dx \]

Before computing, answer:

  1. What is the inner function?
  2. What is its derivative?
  3. Is the derivative present exactly or off by a constant?
  4. What substitution would you choose?

Write your setup before integrating.

13.7.5 Independent Practice

For each integral:

  • Decide whether substitution is appropriate.
  • If yes, state your choice of \(u\) before computing.
  • If no, briefly say why.

  1. \(\displaystyle \int (x^2 + 5)^3 \cdot 2x \, dx\)

  2. \(\displaystyle \int \cos(4x)\, dx\)

  3. \(\displaystyle \int (x^2 + 5)^3 \, dx\)

  4. \(\displaystyle \int \frac{6x}{x^2 - 1} \, dx\)

  5. \(\displaystyle \int x e^x \, dx\)

(At least one of these is not a substitution problem.)

13.7.6 Common Mistakes to Watch For

  • Choosing \(u\) without checking for its derivative
  • Forgetting to convert \(dx\) to \(du\)
  • Substituting back too early
  • Forcing substitution when the structure does not match

If substitution makes the integral messier, that is useful feedback.

13.7.7 Strategy Check

Answer briefly:

  • How do you identify the inner function?
  • When is “off by a constant” acceptable?
  • How can you tell that substitution is not the right method?

Being able to confidently reject substitution is just as important as using it correctly.

13.8 Section 4: Integration by Parts (The Product Rule in Reverse)

Substitution helps when a function is nested inside another function.
Integration by parts is used when an integrand is a product of two functions.

Conceptually, this technique is not new mathematics.
It is the product rule, rearranged to answer a different question.

13.8.1 What to Notice

  • Integration by parts applies when the integrand is a product
  • One factor should become simpler when differentiated
  • The goal is to trade one product for another that is easier to integrate
  • Not every product uses integration by parts
  • If substitution fits the structure, use substitution instead

13.8.2 The Core Idea: The Product Rule in Reverse

Recall the product rule from differentiation: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]

Rearrange this equation to isolate one product: \[ u(x)v'(x) = \frac{d}{dx}[u(x)v(x)] - u'(x)v(x) \]

Now integrate both sides: \[ \int u(x)v'(x)\,dx = u(x)v(x) - \int u'(x)v(x)\,dx \]

This is the integration by parts formula: \[ \int u\,dv = uv - \int v\,du \]

This formula is not meant to be memorized in isolation.
It encodes the product rule.

13.8.3 What the Formula Is Really Doing

Integration by parts moves complexity:

  • one factor is differentiated (often simplifying),
  • the other is integrated (often staying manageable),
  • the remaining integral is simpler than the original.

If the remaining integral is not simpler, the method has failed—and that is useful information.

13.8.4 When Integration by Parts Is a Good Choice

Ask the diagnostic question:

If I differentiate one factor, does it become simpler?

Common situations where the answer is yes:

  • polynomials multiplied by exponentials or trigonometric functions
  • logarithms multiplied by anything
  • inverse trigonometric functions multiplied by algebraic factors

13.8.5 Worked Example (Instructor-Led)

Evaluate: \[ \int x e^x \, dx \]

Pause & Predict

Before computing:

  • Is this a product?
  • Would substitution simplify this?
  • Which factor becomes simpler when differentiated?

Choose \(u\) and \(dv\)

Let: \[ u = x \quad \Rightarrow \quad du = dx \]

Let: \[ dv = e^x\,dx \quad \Rightarrow \quad v = e^x \]

Apply the Formula

\[ \int u\,dv = uv - \int v\,du \]

\[ \int x e^x dx = x e^x - \int e^x dx \]

Finish the Integral

\[ x e^x - e^x + C \]

Or, factored: \[ e^x(x - 1) + C \]

13.8.6 Why This Worked

Differentiating \(x\) reduced the product to a constant.
The remaining integral was simpler than the original.

This is the goal of integration by parts.

13.8.7 A “Bad Choice” Example

Consider the same integral: \[ \int x e^x dx \]

If we choose: \[ u = e^x \]

Then: \[ du = e^x dx \]

The resulting integral remains a product of two nontrivial functions.
Nothing has simplified.

This is a signal to change strategy.

13.8.8 Guided Practice (With Prompts)

Evaluate: \[ \int x \sin x \, dx \]

Before computing, answer:

  1. Which factor becomes simpler when differentiated?
  2. What will you choose for \(u\)?
  3. What will you choose for \(dv\)?

Write your choices before applying the formula.

13.8.9 Independent Practice

For each integral:

  • Decide whether integration by parts is appropriate.
  • If yes, state your choices of \(u\) and \(dv\) before computing.
  • If no, briefly explain why.

  1. \(\displaystyle \int x \cos x \, dx\)

  2. \(\displaystyle \int \ln x \, dx\)

  3. \(\displaystyle \int x^2 e^x \, dx\)

  4. \(\displaystyle \int e^{x^2} \cdot 2x \, dx\)

  5. \(\displaystyle \int (x^2 + 1)^3 \, dx\)

(At least one of these should not use integration by parts.)

13.8.10 Common Mistakes to Watch For

  • Choosing \(u\) that becomes more complicated when differentiated
  • Forgetting the minus sign in the formula
  • Stopping before the integral is fully simplified
  • Using integration by parts when substitution applies

13.8.11 Strategy Check

Answer briefly:

  • How do you decide which factor to differentiate?
  • What does it mean for an integral to become “simpler”?
  • How can a failed attempt guide your next method choice?

Integration by parts is powerful when used deliberately.
When used mechanically, it often makes problems worse.

Learning to recognize when—and why—it works is the real goal.

13.9 Section 5: Partial Fractions (Breaking Rational Functions Apart)

Substitution and integration by parts focus on structure (nesting or products).
Partial fractions is different: it applies to rational functions.

A rational function is a ratio of polynomials.

This technique works by rewriting a complicated fraction as a sum of simpler fractions that are easy to integrate.

13.9.1 What to Notice

  • Partial fractions applies only to rational functions
  • The denominator must be factorable
  • The degree of the numerator must be less than the degree of the denominator
    (otherwise, simplify first)
  • Each factor in the denominator contributes its own term
  • Once decomposed, integration uses only basic rules

13.9.2 The Core Idea: Decomposition Instead of Substitution

The goal is not to “solve” the integral directly.

Instead, you:

  1. Rewrite the integrand as a sum of simpler rational pieces
  2. Integrate each piece using basic logarithm or power rules

Nothing new happens at the integration step.
All the work is in the algebra.

13.9.3 Types of Denominator Factors (In This Course)

In this course, we focus on linear factors only.

13.9.3.1 1. Distinct Linear Factors

Example: \[ \frac{1}{(x-1)(x+2)} \]

Decomposes as: \[ \frac{A}{x-1} + \frac{B}{x+2} \]

Each linear factor gets one term.

13.9.3.2 2. Repeated Linear Factors

Example: \[ \frac{1}{(x-1)^2} \]

Decomposes as: \[ \frac{A}{x-1} + \frac{B}{(x-1)^2} \]

Each power of a repeated factor gets its own term.

13.9.4 Types of Denominator Factors (In This Course)

In this course, we focus on linear factors only.

13.9.4.1 1. Distinct Linear Factors

Example: \[ \frac{1}{(x-1)(x+2)} \]

Decomposes as: \[ \frac{A}{x-1} + \frac{B}{x+2} \]

Each linear factor gets one term.

Integration Form

Once decomposed, each term integrates using the logarithm rule: \[ \int \frac{A}{x+a}\,dx = A \ln|x+a| + C \]

The integration step is straightforward.
All of the difficulty happens before this, during decomposition.

13.9.4.2 2. Repeated Linear Factors

Example: \[ \frac{1}{(x-1)^2} \]

Decomposes as: \[ \frac{A}{x-1} + \frac{B}{(x-1)^2} \]

Each power of a repeated factor gets its own term.

13.9.4.3 Integration Forms

Each term integrates using basic power rules: \[ \int \frac{A}{x+a}\,dx = A \ln|x+a| + C \]

\[ \int \frac{B}{(x+a)^2}\,dx = -\frac{B}{x+a} + C \]

Notice that: - the first term produces a logarithm, - the second produces a rational function, - no substitution is required.

Key Structural Takeaway

  • Distinct linear factors → logarithms
  • Repeated linear factors → logarithms and power-rule terms
  • Partial fractions is an algebra-first technique

If the denominator structure is wrong, the method does not apply.

13.9.5 Out of Scope (For This Course)

You may see references to:

  • irreducible quadratic factors
  • quadratic terms in the numerator

These require additional decomposition rules and lead to arctangent-type integrals.

They exist, but they are out of scope for this course.
Our focus is on structure, not exhaustive case handling.

13.9.6 When Partial Fractions Is a Good Choice

Ask the diagnostic question:

Is this a rational function with a factorable denominator?

If yes:

  • and the denominator factors into linear terms,
  • and the numerator degree is smaller,

then partial fractions is likely the right tool.

If the integrand is not a rational function, partial fractions does not apply.

13.9.7 Worked Example (Instructor-Led)

Evaluate: \[ \int \frac{7x+1}{x^2+x-2}\,dx \]

Pause & Predict

Before computing:

  • Is this a rational function?
  • Does the denominator factor?
  • Are the factors linear and distinct?

Factor the denominator \[ x^2+x-2 = (x-1)(x+2) \]

Set Up the Decomposition

Write: \[ \frac{7x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \]

Solve for \(A\) and \(B\), then integrate each term using basic rules.

13.9.8 Worked Example (Instructor-Led)

Evaluate: \[ \int \frac{5x+3}{(x^2-1}\,dx \]

Pause & Predict

Before computing:

  • Is this a rational function?
  • Does the denominator factor?
  • Is the factor repeated?

Factor the denominator \[ x^2-1 = (x-1)^2 \]

Set Up the Decomposition

Write: \[ \frac{5x+3}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} \]

Each power of the repeated linear factor requires its own term.

Solve for \(A\) and \(B\), then integrate each term using basic rules.

13.9.9 Your Turn

Evaluate each integral using partial fractions.

  1. \[ \int \frac{3}{x^2 - x - 2}\,dx \]
  1. \[ \int \frac{5x-1}{x^2 + 2x - 3}\,dx \]
  1. \[ \int \frac{4}{x^2 - 4x + 4}\,dx \]
  1. \[ \int \frac{6x+1}{x^2 + 2x + 1}\,dx \]
  1. \[ \int \frac{2x+5}{x^2 + x - 6}\,dx \]
  1. \[ \int \frac{7x-3}{x^2 - 5x + 6}\,dx \]
Click to reveal answer key

Answer Key

Solution 1 \[ \int \frac{3}{x^2-x-2}\,dx \]

Step 1: Factor the denominator \[ x^2-x-2=(x-2)(x+1) \]

Step 2: Set up partial fractions \[ \frac{3}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1} \]

Step 3: Clear denominators \[ 3=A(x+1)+B(x-2) \]

Step 4: Solve for \(A,B\)

Plug \(x=2\): \[ 3=A(3)\Rightarrow A=1 \]

Plug \(x=-1\): \[ 3=B(-3)\Rightarrow B=-1 \]

So: \[ \frac{3}{(x-2)(x+1)}=\frac{1}{x-2}-\frac{1}{x+1} \]

Step 5: Integrate \[ \int\left(\frac{1}{x-2}-\frac{1}{x+1}\right)\,dx =\ln|x-2|-\ln|x+1|+C \]

Solution 2 \[ \int \frac{5x-1}{x^2+2x-3}\,dx \]

Step 1: Factor the denominator \[ x^2+2x-3=(x+3)(x-1) \]

Step 2: Set up partial fractions \[ \frac{5x-1}{(x+3)(x-1)}=\frac{A}{x+3}+\frac{B}{x-1} \]

Step 3: Clear denominators \[ 5x-1=A(x-1)+B(x+3) \]

Step 4: Solve for \(A,B\)

Plug \(x=1\): \[ 5(1)-1=B(4)\Rightarrow 4=4B\Rightarrow B=1 \]

Plug \(x=-3\): \[ 5(-3)-1=A(-4)\Rightarrow -16=-4A\Rightarrow A=4 \]

So: \[ \frac{5x-1}{(x+3)(x-1)}=\frac{4}{x+3}+\frac{1}{x-1} \]

Step 5: Integrate \[ \int\left(\frac{4}{x+3}+\frac{1}{x-1}\right)\,dx =4\ln|x+3|+\ln|x-1|+C \]

Solution 3 \[ \int \frac{4}{x^2-4x+4}\,dx \]

Step 1: Recognize the repeated root \[ x^2-4x+4=(x-2)^2 \]

Step 2: Rewrite the integrand \[ \int \frac{4}{(x-2)^2}\,dx = \int 4(x-2)^{-2}\,dx \]

Step 3: Integrate (power rule) \[ 4\cdot \frac{(x-2)^{-1}}{-1}+C = -\frac{4}{x-2}+C \]

Solution 4 \[ \int \frac{6x+1}{x^2+2x+1}\,dx \]

Step 1: Recognize the repeated root \[ x^2+2x+1=(x+1)^2 \]

Step 2: Set up the repeated-root form \[ \frac{6x+1}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2} \]

Step 3: Clear denominators \[ 6x+1=A(x+1)+B \]

Step 4: Solve for \(A,B\)

Match coefficients: \[ 6x+1=Ax+(A+B) \] So: \[ A=6,\quad A+B=1\Rightarrow 6+B=1\Rightarrow B=-5 \]

Thus: \[ \frac{6x+1}{(x+1)^2}=\frac{6}{x+1}-\frac{5}{(x+1)^2} \]

Step 5: Integrate \[ \int \frac{6}{x+1}\,dx=6\ln|x+1| \] \[ \int -\frac{5}{(x+1)^2}\,dx = -5\int (x+1)^{-2}\,dx = -5\left(\frac{(x+1)^{-1}}{-1}\right) = \frac{5}{x+1} \]

Final: \[ 6\ln|x+1|+\frac{5}{x+1}+C \]

Solution 5 \[ \int \frac{2x+5}{x^2+x-6}\,dx \]

Step 1: Factor the denominator \[ x^2+x-6=(x+3)(x-2) \]

Step 2: Set up partial fractions (A and B only) \[ \frac{2x+5}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2} \]

Step 3: Clear denominators \[ 2x+5=A(x-2)+B(x+3) \]

Step 4: Solve for \(A,B\)

Plug \(x=2\): \[ 2(2)+5=B(5)\Rightarrow 9=5B\Rightarrow B=\frac{9}{5} \]

Plug \(x=-3\): \[ 2(-3)+5=A(-5)\Rightarrow -1=-5A\Rightarrow A=\frac{1}{5} \]

So: \[ \frac{2x+5}{(x+3)(x-2)}=\frac{1/5}{x+3}+\frac{9/5}{x-2} \]

Step 5: Integrate \[ \int\left(\frac{1/5}{x+3}+\frac{9/5}{x-2}\right)\,dx = \frac{1}{5}\ln|x+3|+\frac{9}{5}\ln|x-2|+C \]

Solution 6 \[ \int \frac{7x-3}{x^2-5x+6}\,dx \]

Step 1: Factor the denominator \[ x^2-5x+6=(x-2)(x-3) \]

Step 2: Set up partial fractions (A and B only) \[ \frac{7x-3}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3} \]

Step 3: Clear denominators \[ 7x-3=A(x-3)+B(x-2) \]

Step 4: Solve for \(A,B\)

Plug \(x=2\): \[ 7(2)-3=A(-1)\Rightarrow 11=-A\Rightarrow A=-11 \]

Plug \(x=3\): \[ 7(3)-3=B(1)\Rightarrow 18=B \]

So: \[ \frac{7x-3}{(x-2)(x-3)}=\frac{-11}{x-2}+\frac{18}{x-3} \]

Step 5: Integrate \[ \int\left(\frac{-11}{x-2}+\frac{18}{x-3}\right)\,dx = -11\ln|x-2|+18\ln|x-3|+C \]

13.10 Test 2 Review

Practice: Identifying the Appropriate Technique

For each integral below, choose the single most appropriate method.
You do not need to evaluate the integral.

Options:

  • A. Basic integration rules only
  • B. Substitution
  • C. Integration by parts
  • D. Partial fractions

  1. \[ \int (4x^2+1)^6 \, dx \]

  2. \[ \int x e^{x^2+3}\, dx \]

  3. \[ \int \frac{1}{x^2+4x+4}\, dx \]

  4. \[ \int \frac{6x}{x^2+9}\, dx \]

  5. \[ \int x^2 e^{x}\, dx \]

  6. \[ \int \frac{5}{x(x-3)}\, dx \]

  7. \[ \int (x^3+1)\ln x\, dx \]

  8. \[ \int e^{5x}\, dx \]

  9. \[ \int \frac{2x+1}{(x+1)^2}\, dx \]

  10. \[ \int x\cos x\, dx \]

  11. \[ \int \frac{4}{x^2-1}\, dx \]

  12. \[ \int (3x-2)e^{(3x-2)^2}\, dx \]

  13. \[ \int \ln x \, dx \]

  14. \[ \int \frac{x}{\sqrt{x^2-4}}\, dx \]

  15. \[ \int x^3\sin x\, dx \]

  16. \[ \int (x^2-5)^4\, dx \]

  17. \[ \int \frac{3}{(x-2)(x+5)}\, dx \]

  18. \[ \int x e^{-x}\, dx \]

Evaluate each integral using substitution

1. \[ \int 4x(2x^2-3)^6\,dx \]

2. \[ \int \frac{7x}{x^2+1}\,dx \]

3. \[ \int e^{5x+2}\,dx \]

4. \[ \int \frac{3x}{\sqrt{x^2-4}}\,dx \]

5. \[ \int (4x-1)^5\,dx \]

Practice: Integration by Parts

Evaluate each integral using integration by parts.

1. \[ \int x e^{-3x}\,dx \]

2. \[ \int x^3 e^{x}\,dx \]

3. \[ \int x\ln(2x)\,dx \] Hint: \[ \frac{d (ln(g(x))}{dx} = \frac{g'(x))}{g(x)}\]

4. \[ \int (x^2+3x)\sin x\,dx \]

Hint: using Linearity might help keep things organized

5. \[ \int x^{-2}\ln(x)\,dx = \int \frac{\ln x}{x^2}\,dx. \]

Practice: Partial Fractions (Factorable Quadratics)

Evaluate each integral using partial fractions.
All denominators are quadratic and factorable (repeated factors may appear).

1. \[ \int \frac{8}{x^2-16}\,dx \]

2. \[ \int \frac{5x}{x^2-3x}\,dx \]

3. \[ \int \frac{2x-7}{x^2-9x+18}\,dx \]

4. \[ \int \frac{6x+2}{x^2+2x+1}\,dx \]

5. \[ \int \frac{x+3}{(x+1)(x+3)}\,dx \]

Hint: Can you see a shortcut?

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