courses › Calculus

Integration techniques

level 44 course

Substitution, integration by parts, and improper integrals.

Pen and paper is fine · no calculator needed why?

Opens at level 44.

Builds on: Integrals (not open yet)

the lesson

The idea, the techniques and a tip for each skill, right here. The Learn page adds worked examples for every skill and untimed practice.

Read the lesson · about 3 minutes

The idea

Some integrals need rearranging before the power rule can finish them.

Substitution undoes the chain rule: when the integrand holds an inside function and that inside's derivative, rename the inside u. Integration by parts undoes the product rule: ∫ u dv = uv − ∫ v du trades a product for something simpler.

An improper integral runs to infinity; the antiderivative's value as x grows stands in for the top limit.

Techniques

Substitute, and change the limits

An inside function times its derivative, or a multiple of it, like 2x(x² + 1)³ or x(x² + 3)².

Let u be the inside, and find du: for u = x² + 1, du = 2x dx.
If the front factor is short, divide by the missing number: x dx = du/2.
Change the limits: put each x-limit into u.
Integrate in u between the new limits: uⁿ becomes uⁿ⁺¹/(n + 1).

worked example

Example: What is ∫₀¹ 3x²(x³ + 1)² dx? Enter a fraction or a whole number.

Let u = x³ + 1, so du = 3x² dx.
x = 0 gives u = 1. x = 1 gives u = 2.
Integrate u² from 1 to 2: (8 − 1)/3 = 7/3.

Answer: 7/3

Integration by parts

x·eˣ, x·sin(x) or x·cos(x), or ln(x) on its own.

Let u be the part that gets simpler when differentiated, x or ln(x). Let dv be the rest.
Find du and v, and use ∫ u dv = uv − ∫ v du. Then take top limit minus bottom.
Useful values: sin π = 0, cos π = −1, sin(π/2) = 1, ln(e) = 1, ln(1) = 0.

worked example

Example: What is ∫ x·cos(x) dx from x = 0 to x = π?

u = x and dv = cos(x) dx, so du = dx and v = sin(x).
Integral: x·sin(x) − ∫ sin(x) dx = x·sin(x) + cos(x).
At π: 0 + (−1) = −1. At 0: 0 + 1 = 1. So −1 − 1 = −2.

Answer: −2

Improper integrals to infinity

A top limit of ∞, with A/xᵖ (p above 1) or e^(−kx) (k above 0).

Find the antiderivative: A/xᵖ becomes −A/((p − 1)xᵖ⁻¹), and e^(−kx) becomes −e^(−kx)/k.
As x → ∞, both of those shrink to 0.
Answer: 0 minus the antiderivative at the lower limit.

worked example

Example: What is ∫₁^∞ 6/x⁴ dx? Enter a fraction or a whole number.

6/x⁴ is 6x⁻⁴, so an antiderivative is −2/x³.
As x → ∞ that goes to 0. At x = 1 it is −2.
0 − (−2) = 2.

Answer: 2

Tips by skill

TipSubstitution: Let u be the inside and find du. Change the limits into u, and watch for a leftover factor like 1/2.
TipIntegration by parts: Let u be x (or ln x) and dv the rest. Use uv − ∫ v du, then evaluate at both limits.
TipImproper integrals: The antiderivative goes to 0 as x → ∞. The answer is 0 minus its value at the lower limit.

Watch out for

Switching to u but keeping the old x-limits. Put each limit into u first.
Stopping after the uv part. The − ∫ v du part still has to be done.
Sign slips: the antiderivative of sin(x) is −cos(x), and it is always top minus bottom.

skills · practice stats

From rounds of this course only: box, review and test-out answers are left out. Once a skill has 40 tries, it compares your first 20 tries with your last 20.

Substitution par 45 s not tried yet
worked example
What is ∫₀¹ 2x(x² + 2)³ dx? Enter a fraction or a whole number.

Answer: 65/4
1. Let u = x² + 2; then du = 2x dx. x = 0 gives u = 2 and x = 1 gives u = 3.
2. The integral becomes ∫ u³ du from u = 2 to u = 3.
3. That is (3⁴ − 2⁴)/4 = (81 − 16)/4 = 65/4.
Integration by parts par 60 s not tried yet
worked example
What is ∫ 6x·cos(x) dx from x = 0 to x = π?

Answer: -12
1. By parts with u = x and dv = cos(x) dx: ∫ x·cos(x) dx = x·sin(x) + cos(x).
2. At π: 0 + (−1) = −1. At 0: 0 + 1 = 1.
3. So the integral is 6 × (−1 − 1) = −12.
Improper integrals par 40 s not tried yet
worked example
What is ∫₁^∞ 9/x³ dx? Enter a fraction or a whole number.

Answer: 9/2
1. 9/x³ = 9x⁻³, so an antiderivative is 9x⁻²/(−2) = −9/(2x²).
2. As x → ∞ it goes to 0; at x = 1 it is −9/2.
3. So the integral is 0 − (−9/2) = 9/2.

rest ladder

Win 3 of your last 4 rounds and the course rests. A win is 90% right, within 2× the round's par. Pass the review when it comes back and the next rest is longer.

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