Intermediate Integration Problems- Challenging Practice with Solutions

What Makes Integration Problems "Intermediate"?

Not all integration problems are created equal. At the beginner level, you're mostly applying basic formulas: power rules, simple trig, straightforward substitutions. Intermediate problems require you to combine multiple techniques in a single problem, recognize hidden patterns, and know when one method beats another.

If you've mastered basic antiderivatives but hit a wall when problems get messy, this guide is for you. We'll walk through the main techniques, compare them clearly, and solve real problems step by step.

The Core Techniques You Need to Know

Before diving into problems, let's establish what arsenal you're working with. Intermediate integration is really about choosing the right tool and executing it cleanly.

1. Substitution (U-Substitution)

The workhorse of integration. Use it when you spot a function and its derivative present in the integrand. The trick is identifying what to set equal to u.

2. Integration by Parts

When substitution fails and you have products of different function types (polynomial × exponential, logarithm × polynomial), integration by parts often works. Remember: LIATE — pick u based on this order: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.

3. Trigonometric Substitution

Essential when you see √(a² - x²), √(a² + x²), or √(x² - a²) in the integrand. Each pattern corresponds to a specific substitution using sine, tangent, or secant.

4. Partial Fractions

For rational functions where the numerator's degree is lower than the denominator's, decompose the fraction into simpler terms. This is mechanical once you understand the decomposition patterns.

5. Trigonometric Integration

Powers of sine and cosine require specific strategies. Use power-reducing identities, or apply the half-angle formulas strategically based on whether powers are odd or even.

Comparing Integration Techniques

Getting Started: A Systematic Approach

Here's how to tackle any intermediate integration problem:

1. Look before you substitute. Scan the entire integrand. Is there a pattern? A derivative hiding? A factorization that simplifies things?
2. Try substitution first. It's often the simplest path. Let u equal something that simplifies the integrand.
3. Check if parts is needed. If you have a product and substitution doesn't work, integration by parts is your next move.
4. Watch for special forms. Trig substitutions and partial fractions have obvious triggers — don't force other methods when these patterns appear.
5. Simplify ruthlessly. Factor, cancel, rewrite — do whatever makes the integral cleaner before committing to a technique.

Problem 1: U-Substitution with a Twist

Solve: ∫ x³√(x² + 1) dx

At first glance, substitution looks straightforward. But the x³ doesn't match the derivative of x² + 1 (which is 2x). We need to split the x³.

Let u = x² + 1. Then du = 2x dx, so x dx = du/2.

Rewrite x³ dx as x² · x dx. Since x² = u - 1:

∫ x³√(x² + 1) dx = ∫ (u - 1)√u · (du/2)

= (1/2) ∫ (u - 1)u^(1/2) du = (1/2) ∫ (u^(3/2) - u^(1/2)) du

Integrate:

= (1/2) [ (2/5)u^(5/2) - (2/3)u^(3/2) ] + C

= (1/5)u^(5/2) - (1/3)u^(3/2) + C

Substitute back:

= (1/5)(x² + 1)^(5/2) - (1/3)(x² + 1)^(3/2) + C

Problem 2: Integration by Parts — When It Hurts

Solve: ∫ x² ln(x) dx

This screams parts. Ln(x) has no elementary antiderivative, so it must be u. The algebraic x² becomes dv.

Set u = ln(x), dv = x² dx

Then du = (1/x) dx, v = x³/3

Apply the formula: ∫ u dv = uv - ∫ v du

= (x³/3)ln(x) - ∫ (x³/3)(1/x) dx

= (x³/3)ln(x) - (1/3)∫ x² dx

= (x³/3)ln(x) - (1/3)(x³/3) + C

= (x³/3)ln(x) - x³/9 + C

Problem 3: Trigonometric Substitution

Solve: ∫ dx / (x²√(x² + 4))

The √(x² + 4) is the trigger. This fits the √(x² + a²) pattern, so we use x = 2tan(θ).

Then dx = 2sec²(θ) dθ, and √(x² + 4) = √(4tan²θ + 4) = 2sec(θ)

Also x² = 4tan²(θ)

Substitute:

∫ [2sec²(θ) dθ] / [4tan²(θ) · 2sec(θ)] = ∫ [2sec²(θ)] / [8tan²(θ)sec(θ)] dθ

= (1/4) ∫ sec(θ) / tan²(θ) dθ

Rewrite: sec(θ)/tan²(θ) = (1/cosθ) · (cos²θ/sin²θ) = cosθ/sin²θ

= (1/4) ∫ cosθ/sin²θ dθ

Let u = sin(θ), du = cosθ dθ

= (1/4) ∫ du/u² = (1/4)(-1/u) + C = -1/(4sinθ) + C

Now convert back. Since x = 2tanθ, we have tanθ = x/2, so sinθ = x/√(x² + 4)

= -√(x² + 4) / (4x) + C

Problem 4: Partial Fractions

Solve: ∫ (3x + 2) / (x² - x - 2) dx

First factor the denominator: x² - x - 2 = (x - 2)(x + 1)

Set up partial fractions:

(3x + 2) / [(x - 2)(x + 1)] = A/(x - 2) + B/(x + 1)

Multiply both sides by (x - 2)(x + 1):

3x + 2 = A(x + 1) + B(x - 2)

Solve for A and B. Set x = 2: 3(2) + 2 = A(3), so 8 = 3A, A = 8/3

Set x = -1: 3(-1) + 2 = B(-3), so -1 = -3B, B = 1/3

Rewrite the integral:

∫ [8/3)/(x - 2) + (1/3)/(x + 1)] dx

= (8/3)ln|x - 2| + (1/3)ln|x + 1| + C

= (1/3)[8ln|x - 2| + ln|x + 1|] + C

Problem 5: Trig Power Reduction

Solve: ∫ sin⁴(x) dx

Even power. Use the power-reducing identity:

sin²(x) = (1 - cos(2x))/2

Apply it twice:

sin⁴(x) = [sin²(x)]² = [(1 - cos(2x))/2]² = (1 - 2cos(2x) + cos²(2x))/4

Now reduce cos²(2x) again: cos²(2x) = (1 + cos(4x))/2

Substitute:

sin⁴(x) = [1 - 2cos(2x) + (1 + cos(4x))/2] / 4

= [1 - 2cos(2x) + 1/2 + (1/2)cos(4x)] / 4

= [3/2 - 2cos(2x) + (1/2)cos(4x)] / 4

= 3/8 - (1/2)cos(2x) + (1/8)cos(4x)

Integrate term by term:

∫ sin⁴(x) dx = ∫ [3/8 - (1/2)cos(2x) + (1/8)cos(4x)] dx

= (3x/8) - (1/4)sin(2x) + (1/32)sin(4x) + C

Common Pitfalls to Avoid

• Forcing the wrong method. If substitution keeps making things messier, try parts or partial fractions instead.
• Losing track of constants. Every time you substitute, track your du and adjust coefficients immediately.
• Forgetting to back-substitute. The answer must be in terms of the original variable.
• Partial fraction algebra errors. Always verify your A, B values by plugging test values back in.
• Wrong trig substitution choice. Memorize: a² - x² → sin, a² + x² → tan, x² - a² → sec.

When to Move Forward

If you can work through these five problems without checking solutions first, you've got the basics down. The real skill comes from recognizing which technique applies before you start solving. That recognition comes from practice — not passive reading.

Work through 20-30 more problems. Mix them up. Some will trick you into trying the wrong method first. That's the point. The struggle is where the learning happens.