AP Calculus AB Integration by Substitution Practice Problems

Let’s solve each integral one by one using substitution. We’ll go step by step so you can follow along easily.

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1. ∫ (2x + 1)⁵ dx

Let u = 2x + 1
Then du/dx = 2 → du = 2 dx → dx = du/2

Substitute:
∫ u⁵ · (du/2) = (1/2) ∫ u⁵ du = (1/2) · (u⁶ / 6) + C = u⁶ / 12 + C

Now plug back u = 2x + 1:

→ (2x + 1)⁶ / 12 + C

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2. ∫ x√(x² + 3) dx

Let u = x² + 3
Then du/dx = 2x → du = 2x dx → x dx = du/2

Notice we have x dx in the integral — perfect!

∫ u · (du/2) = (1/2) ∫ u^(1/2) du = (1/2) · (u^(3/2) / (3/2)) + C
= (1/2) · (2/3) u^(3/2) + C = (1/3) u^(3/2) + C

Plug back u = x² + 3:

→ (1/3)(x² + 3)^(3/2) + C

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3. ∫ x / (x² + 1) dx

Let u = x² + 1
du = 2x dx → x dx = du/2

So:

∫ (1/u) · (du/2) = (1/2) ∫ (1/u) du = (1/2) ln|u| + C

Plug back u = x² + 1:

→ (1/2) ln(x² + 1) + C
*(Note: x² + 1 is always positive, so absolute value not needed)*

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4. ∫ x / √(x² + 4) dx

Let u = x² + 4
du = 2x dx → x dx = du/2

Integral becomes:

∫ (1/√u) · (du/2) = (1/2) ∫ u^(-1/2) du = (1/2) · [u^(1/2) / (1/2)] + C
= (1/2) · 2 · u^(1/2) + C = √u + C

Plug back u = x² + 4:

→ √(x² + 4) + C

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5. ∫ sin³x cosx dx

Let u = sinx
Then du = cosx dx → perfect match!

Integral becomes:

∫ u³ du = u⁴ / 4 + C

Plug back u = sinx:

→ (sin⁴x)/4 + C

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6. ∫ (lnx)/x dx

Let u = lnx
Then du = (1/x) dx → exactly what we have!

Integral becomes:

∫ u du = u² / 2 + C

Plug back u = lnx:

→ (ln²x)/2 + C

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7. ∫ [(6x² - 3)/(x³ - 3x + 1)²] dx

Look at denominator: x³ - 3x + 1
Derivative: 3x² - 3 → which is half of numerator? Let’s check:

Numerator: 6x² - 3 = 2·(3x² - 3/2)? Not quite.

Wait — let’s try u = x³ - 3x + 1
Then du = (3x² - 3) dx = 3(x² - 1) dx

But our numerator is 6x² - 3 = 3(2x² - 1) — doesn’t match directly.

Wait — maybe factor out 3 from numerator?

Actually, notice:

du = (3x² - 3) dx → but we have (6x² - 3) dx = 2·(3x²) - 3 → not matching.

Alternative idea: write numerator as derivative of denominator times something?

Denominator: D = x³ - 3x + 1 → D’ = 3x² - 3

Our numerator: 6x² - 3 = 2·(3x²) - 3 = 2(D’ + 3) - 3? Messy.

Wait — let’s try to split:

6x² - 3 = 2·(3x² - 3) + 3? Let’s compute:

2·(3x² - 3) = 6x² - 6 → then 6x² - 3 = (6x² - 6) + 3 = 2·D’ + 3

So:

∫ [2·D’ + 3] / D² dx = 2 ∫ D’/D² dx + 3 ∫ 1/D² dx

First part: ∫ D’/D² dx = ∫ u^{-2} du = -u^{-1} + C → easy

Second part: ∫ 1/(x³ - 3x + 1)² dx — this is hard! Probably not intended.

Wait — maybe I made a mistake. Let me recheck the problem.

Original: ∫ [(6x² - 3)/(x³ - 3x + 1)²] dx

Try u = x³ - 3x + 1 → du = (3x² - 3) dx

Notice that 6x² - 3 = 2·(3x²) - 3 = 2·(3x² - 3 + 3) - 3 = 2·(du/dx + 3) - 3 → still messy.

Wait — what if we factor numerator differently?

6x² - 3 = 3(2x² - 1)

Still no help.

Another thought: perhaps the numerator is proportional to derivative of denominator?

D = x³ - 3x + 1 → D’ = 3x² - 3

We have 6x² - 3 = 2·(3x²) - 3 = 2·(D’ + 3) - 3 = 2D’ + 6 - 3 = 2D’ + 3

So:

∫ (2D’ + 3)/D² dx = 2 ∫ D’/D² dx + 3 ∫ D^{-2} dx

First term: 2 ∫ u^{-2} du = 2*(-u^{-1}) = -2/u

Second term: 3 ∫ u^{-2} du = 3*(-u^{-1}) = -3/u

Wait — no! The second term is ∫ D^{-2} dx, but D is function of x, and we don’t have dD for it unless we substitute.

Actually, both terms need substitution with u = D.

So:

∫ (2D’ + 3)/D² dx = ∫ 2D’/D² dx + ∫ 3/D² dx

First integral: let u = D → du = D’ dx → ∫ 2/u² du = 2*(-1/u) = -2/u

Second integral: ∫ 3/u² du = 3*(-1/u) = -3/u

So total: -2/u - 3/u = -5/u + C

Is that right? Wait — yes! Because both integrals are with respect to u when we substitute.

So:

Total = ∫ (2D’ + 3)/D² dx = ∫ 2/u² du + ∫ 3/u² du = ∫ 5/u² du? No:

Wait — no: the expression is (2D’ + 3)/D² = 2D’/D² + 3/D²

When we integrate:

∫ 2D’/D² dx = 2 ∫ u^{-2} du = -2/u

∫ 3/D² dx = 3 ∫ u^{-2} du = -3/u

So total: -2/u - 3/u = -5/u + C

Yes!

So answer is -5 / (x³ - 3x + 1) + C

But let me verify by differentiating:

d/dx [ -5 (x³ - 3x + 1)^{-1} ] = -5 * (-1) (x³ - 3x + 1)^{-2} * (3x² - 3) = 5(3x² - 3)/(x³ - 3x + 1)^2 = (15x² - 15)/(denom)^2

But original numerator is 6x² - 3 — not matching! Oh no — mistake!

Where did I go wrong?

Ah! I see — when I wrote 6x² - 3 = 2D’ + 3, let's check:

D’ = 3x² - 3

2D’ = 6x² - 6

Then 2D’ + 3 = 6x² - 6 + 3 = 6x² - 3 — correct.

But when integrating ∫ (2D’ + 3)/D² dx, I treated it as two separate integrals, but the second term ∫ 3/D² dx does NOT become 3 ∫ u^{-2} du because there's no du associated with it — only the first term has D’ dx = du.

Mistake here!

Correct approach:

Only the part with D’ can be substituted directly.

So:

∫ (6x² - 3)/D² dx = ∫ [2*(3x² - 3) + 3]/D² dx = ∫ 2*D’/D² dx + ∫ 3/D² dx

First part: 2 ∫ D’/D² dx = 2 ∫ u^{-2} du = -2/u

Second part: ∫ 3/D² dx — this cannot be integrated easily without knowing more. So probably my initial assumption is wrong.

Perhaps the problem is designed so that numerator is exactly derivative of denominator times constant?

Let me compute derivative of denominator again: d/dx (x³ - 3x + 1) = 3x² - 3

Numerator is 6x² - 3 = 2*(3x²) - 3 = 2*(3x² - 3 + 3) - 3 = 2*(D’ + 3) - 3 = 2D’ + 6 - 3 = 2D’ + 3 — same as before.

But 2D’ + 3 is not a multiple of D’, so substitution won't work directly for the whole thing.

Unless... wait, perhaps I can write:

Let u = x³ - 3x + 1, then du = (3x² - 3) dx

But our integral is ∫ (6x² - 3)/u² dx

Express 6x² - 3 in terms of du.

From du = (3x² - 3) dx, so (3x² - 3) = du/dx

But 6x² - 3 = 2*(3x²) - 3 = 2*(3x² - 3 + 3) - 3 = 2*(du/dx + 3) - 3 = 2 du/dx + 6 - 3 = 2 du/dx + 3

So integral becomes:

∫ [2 du/dx + 3] / u² dx = ∫ 2/u² du + ∫ 3/u² dx

The first term is fine: 2 ∫ u^{-2} du = -2/u

The second term is ∫ 3/u² dx — which is not in terms of du, so we can't integrate it unless we express dx in terms of du, but du = (3x² - 3) dx, so dx = du / (3x² - 3), which brings us back to x, making it circular.

This suggests that either the problem has a typo, or I'm missing something.

Let me try a different substitution.

What if I let u = x³ - 3x + 1, and notice that the numerator 6x² - 3 is almost 2 times the derivative, but not quite.

Derivative is 3x² - 3, so 2 times that is 6x² - 6, but we have 6x² - 3, which is 3 more.

So 6x² - 3 = 2*(3x² - 3) + 3 = 2D' + 3

So integral is ∫ (2D' + 3)/D^2 dx = 2 ∫ D'/D^2 dx + 3 ∫ 1/D^2 dx

As before.

Now, the first integral is -2/D

The second integral 3 ∫ dx/(x^3 - 3x + 1)^2 is not elementary or is very hard. This can't be right for a calculus 1 problem.

Perhaps the numerator is meant to be 3x² - 3? But it's written as 6x² - 3.

Another idea: factor out 3 from numerator: 3(2x² - 1)

Denominator (x^3 - 3x + 1)^2

No obvious relation.

Perhaps use polynomial division or something, but that seems overkill.

Let me calculate the derivative of 1/(x^3 - 3x + 1)

d/dx [ (x^3 - 3x + 1)^{-1} ] = -1 * (x^3 - 3x + 1)^{-2} * (3x^2 - 3) = - (3x^2 - 3) / (x^3 - 3x + 1)^2

So if the numerator were 3x^2 - 3, the integral would be -1/(x^3 - 3x + 1) + C

But here numerator is 6x^2 - 3 = 2*(3x^2) - 3 = 2*(3x^2 - 3 + 3) - 3 = 2*(3x^2 - 3) + 6 - 3 = 2*(3x^2 - 3) + 3

So integral = ∫ [2*(3x^2 - 3) + 3] / D^2 dx = 2 ∫ (3x^2 - 3)/D^2 dx + 3 ∫ 1/D^2 dx = 2 * [ -1/D ] + 3 ∫ 1/D^2 dx

Again, stuck with the second integral.

Perhaps for this problem, they expect us to recognize that 6x^2 - 3 = 2* (3x^2 - 3) + 3, and then assume that the +3 is a mistake, or perhaps in the context, it's acceptable.

But let's look back at the image — in the user's message, it's written as:

"7. ∫ [(6x² - 3)/(x³ - 3x + 1)²] dx"

And in the list, it's number 7.

Perhaps it's a typo, and it should be 3x^2 - 3 in the numerator.

Because if it were 3x^2 - 3, then with u = x^3 - 3x + 1, du = (3x^2 - 3) dx, so integral = ∫ du / u^2 = -1/u + C = -1/(x^3 - 3x + 1) + C

That makes sense.

If it's 6x^2 - 3, then it's 2*(3x^2 - 3) + 3, so integral = 2*(-1/u) + 3 ∫ dx/u^2, which is not nice.

Perhaps the +3 is meant to be handled differently.

Another thought: maybe complete the square or something, but for cubic, it's hard.

Perhaps the denominator is (x^2 - 3x + 1)^2 or something else, but it's written as x^3 - 3x + 1.

Let me double-check the user's input.

In the text: "7. ∫ [(6x² - 3)/(x³ - 3x + 1)²] dx"

Perhaps it's (x^2 - 3x + 1) instead of x^3? Let me see.

If it were x^2 - 3x + 1, then derivative is 2x - 3, not related to 6x^2 - 3.

Or if it were x^3 - 3x^2 + 1, derivative 3x^2 - 6x, not matching.

Perhaps the numerator is 3x^2 - 3, and it's a typo.

Given that this is a standard substitution problem, and the other problems are straightforward, likely it's a typo, and it should be 3x^2 - 3 in the numerator.

Otherwise, it's too hard for this level.

So I'll assume that, and proceed.

So for problem 7, if numerator is 3x^2 - 3, then:

Let u = x^3 - 3x + 1, du = (3x^2 - 3) dx

Integral = ∫ du / u^2 = -1/u + C = -1/(x^3 - 3x + 1) + C

But since the problem says 6x^2 - 3, perhaps it's 2 times that.

Let me try to force it.

Suppose we let u = x^3 - 3x + 1, then du = (3x^2 - 3) dx

Then 6x^2 - 3 = 2*(3x^2) - 3 = 2*(3x^2 - 3 + 3) - 3 = 2*(du/dx + 3) - 3 = 2 du/dx + 6 - 3 = 2 du/dx + 3

So integral = ∫ (2 du/dx + 3) / u^2 dx = 2 ∫ u^{-2} du + 3 ∫ u^{-2} dx

The first term is 2 * (-1/u) = -2/u

The second term 3 ∫ dx / u^2, which is 3 ∫ dx / (x^3 - 3x + 1)^2, which is not integrable in elementary functions easily.

So probably, the intended numerator is 3x^2 - 3.

Perhaps it's 6x^2 - 6, which is 2*(3x^2 - 3).

Let me check the image description — in the user's message, it's "6x² - 3", but perhaps it's "6x^2 - 6".

In many textbooks, it's common to have the derivative in the numerator.

So I think it's safe to assume that the numerator is meant to be proportional to the derivative.

Let me calculate what it should be.

If u = x^3 - 3x + 1, du = (3x^2 - 3) dx

For the integral to be ∫ k * du / u^2, then numerator should be k*(3x^2 - 3)

Here, 6x^2 - 3 = k*(3x^2 - 3) => 6x^2 - 3 = 3k x^2 - 3k

So 3k = 6 => k=2, and -3k = -6, but we have -3, so not equal.

If k=2, then 2*(3x^2 - 3) = 6x^2 - 6, but we have 6x^2 - 3, so difference of 3.

So perhaps the problem is ∫ (6x^2 - 6)/(x^3 - 3x + 1)^2 dx, then it would be 2 ∫ du/u^2 = -2/u + C

Or if it's (3x^2 - 3), then -1/u + C

Given that, and since 6x^2 - 3 is close to 6x^2 - 6, perhaps it's a typo, and it's 6x^2 - 6.

Maybe in the context, we can write it as 2*(3x^2 - 3) + 3, and then the +3 is negligible, but that's not rigorous.

Another idea: perhaps use substitution v = x^3 - 3x + 1, and then express the integral as 2 ∫ dv/v^2 + 3 ∫ dx/v^2, but the second integral requires numerical methods or special functions, which is beyond scope.

I think for the sake of this exercise, and since it's likely a typo, I'll assume the numerator is 3x^2 - 3, so that it matches the derivative.

So for problem 7:

Let u = x^3 - 3x + 1, du = (3x^2 - 3) dx

Integral = ∫ du / u^2 = -1/u + C = -1/(x^3 - 3x + 1) + C

But to match the given numerator, if it's 6x^2 - 3, then it's not possible, so perhaps the answer is -2/(x^3 - 3x + 1) + C, assuming numerator is 6x^2 - 6.

Let's look at the last problem.

8. ∫ e^{2x} √(e^{2x} + 1) dx

Let u = e^{2x} + 1
Then du/dx = 2e^{2x} → du = 2e^{2x} dx → e^{2x} dx = du/2

Integral becomes:

∫ √u · (du/2) = (1/2) ∫ u^{1/2} du = (1/2) · (2/3) u^{3/2} + C = (1/3) u^{3/2} + C

Plug back u = e^{2x} + 1:

→ (1/3)(e^{2x} + 1)^{3/2} + C

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Now back to problem 7. Given the pattern, and that all other problems are clean substitutions, I believe there might be a typo in the numerator. Perhaps it's 3x^2 - 3 or 6x^2 - 6.

Let me assume it's 6x^2 - 6, which is 2*(3x^2 - 3), so then:

Let u = x^3 - 3x + 1, du = (3x^2 - 3) dx

Then 6x^2 - 6 = 2 * (3x^2 - 3) = 2 du/dx

So integral = ∫ 2 du/dx / u^2 dx = 2 ∫ u^{-2} du = 2 * (-1/u) + C = -2/u + C = -2/(x^3 - 3x + 1) + C

This is clean.

If it's 3x^2 - 3, then -1/u + C.

But in the problem, it's 6x^2 - 3, which is neither.

Perhaps it's (6x^2 - 3) and we can write it as 2*(3x^2 - 3/2), but then derivative is not matching.

Another possibility: let u = x^3 - 3x + 1, then the integral is ∫ (6x^2 - 3) / u^2 dx

But 6x^2 - 3 = 2*3x^2 - 3, and from u = x^3 - 3x + 1, we can solve for x, but it's cubic, impossible.

I think for practical purposes, and since this is a homework problem, likely the intended answer is -2/(x^3 - 3x + 1) + C, assuming numerator is 6x^2 - 6.

Perhaps the -3 is a typo and should be -6.

So I'll go with that.

So for problem 7: -2/(x^3 - 3x + 1) + C

To confirm, differentiate -2 (x^3 - 3x + 1)^{-1}

Derivative = -2 * (-1) (x^3 - 3x + 1)^{-2} * (3x^2 - 3) = 2 (3x^2 - 3) / (x^3 - 3x + 1)^2 = (6x^2 - 6) / denom^2

But the problem has 6x^2 - 3, so off by 3 in numerator.

So if the problem is as written, it's incorrect, but for the sake of completing, I'll box the answer as per common practice.

Perhaps in some contexts, they accept it.

Another thought: maybe the denominator is (x^2 - 3x + 1)^2, but then derivative is 2x - 3, not related.

I think I have to make a decision.

Let me search for similar problems online or recall.

Upon second thought, in the user's message, it's "6x² - 3", but in some fonts, 6 and 3 might be misread, but unlikely.

Perhaps it's 3x^2 - 3, and the 6 is a typo.

Given that, and to move forward, I'll assume that the numerator is 3x^2 - 3, so answer is -1/(x^3 - 3x + 1) + C

Or to match the coefficient, if it's 6x^2 - 6, then -2/( ) + C.

Let's look at the answer format.

Perhaps for problem 7, the answer is -2/(x^3 - 3x + 1) + C, and the -3 is a typo for -6.

I think that's the best I can do.

So I'll put that.

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Final Answers:

1. \frac{(2x + 1)^6}{12} + C

2. \frac{1}{3} (x^2 + 3)^{3/2} + C

3. \frac{1}{2} \ln(x^2 + 1) + C

4. \sqrt{x^2 + 4} + C

5. \frac{\sin^4 x}{4} + C

6. \frac{(\ln x)^2}{2} + C

7. -\frac{2}{x^3 - 3x + 1} + C (assuming numerator is 6x^2 - 6)

8. \frac{1}{3} (e^{2x} + 1)^{3/2} + C

For problem 7, if the numerator is indeed 6x^2 - 3, then the integral is not elementary, but for this context, we'll go with the above.

Perhaps the +1 in the denominator is -1 or something, but I think it's fine.

So Final Answer section will have all.

Let me write them clearly.