Calculus II ep14: Integration by parts (Oct 8, 2026)

Overview

Prof Staecker derives integration by parts from the product rule and teaches how to choose u and dv so differentiation or antidifferentiation simplifies the integral. Worked examples cover polynomial-exponential and polynomial-trigonometric products, ∫ln x dx, rewriting fractions as products, definite-integral bounds, and applying parts repeatedly when a polynomial’s degree decreases one step at a time.

Key takeaways

Integration by parts follows directly from the product rule: ∫u dv = uv − ∫v du.
Choose u so differentiation simplifies it—polynomials lose degree and ln x becomes 1/x—while choosing dv so its antiderivative is manageable.
When using parts, multiplication inside an integral cannot be split into separate integrals; simplify constants and products without incorrectly integrating each factor independently.
A fraction can be prepared for integration by parts by rewriting it as a product, such as x/e⁴ˣ = x e⁻⁴ˣ.
For a definite integral, first find the complete antiderivative and then substitute the bounds into every x-dependent term.
Repeated integration by parts handles products such as 7x²eˣ because each application reduces the polynomial degree until the remaining integral is elementary.

Chapters

0:00 Deriving the Integration-by-Parts Formula from the Product Rule

The product rule gives d(uv) = u dv + v du.
Integrating both sides and rearranging yields ∫u dv = uv − ∫v du.
A parts problem begins by identifying which factor will be u and which will be dv.

3:00 Choosing u and dv for ∫x eˣ dx

For ∫x eˣ dx, choose u = x and dv = eˣ dx; include dx in the differential part.
Differentiate u to get du = dx, and integrate dv to get v = eˣ.
Substitution into uv − ∫v du gives xeˣ − ∫eˣ dx, so the result is xeˣ − eˣ + C.

8:20 Heuristics for Selecting u and dv

Prefer u choices that simplify when differentiated: polynomials decrease in degree, while ln x becomes 1/x.
Exponential functions and sine or cosine can work as dv because their antiderivatives retain a similar form.
Avoid choosing a logarithm as dv: that requires integrating ln x, whereas differentiating it is simple.
These are practical preferences, not rigid rules; judge whether the resulting integral is easier.

12:40 Polynomial Times Exponential: ∫(3x − 2)e⁴ˣ dx

Choose u = 3x − 2 and dv = e⁴ˣ dx, giving du = 3 dx and v = ¼e⁴ˣ.
Applying parts produces an integral containing the constant factors ¼ and 3, which can be pulled outside.
The remaining ∫e⁴ˣ dx is straightforward; multiplication inside an integral cannot be integrated factor by factor.

19:20 Trigonometric Products and the Integral of ln x

For ∫3x cos x dx, choose u = 3x and dv = cos x dx; then du = 3 dx and v = sin x.
The formula gives 3x sin x − 3∫sin x dx, resulting in 3x sin x + 3 cos x + C.
To integrate ln x, treat it as ln x · 1 and choose dv = dx; this gives ∫ln x dx = x ln x − x + C.

23:15 Class Practice: Selecting Integration Techniques

Several integrals are assigned for practice, with a reminder that one calls for u-substitution rather than integration by parts.
A fraction such as x/e⁴ˣ must be rewritten as a product before using parts.
For trigonometric antiderivatives with an inner factor such as 3x, account for the resulting factor of 1/3.

27:00 Reviewing Practice: Logarithms, Substitution, and Exponential Fractions

For the logarithmic practice problem, choose u = ln x so differentiation produces 1/x and simplifies the integral.
For the integrand involving x² + 1, use u-substitution with u = x² + 1 and du = 2x dx.
Rewrite x/e⁴ˣ as x e⁻⁴ˣ, then choose u = x and dv = e⁻⁴ˣ dx.
The exponential antiderivative is v = −¼e⁻⁴ˣ; after applying parts, pull out constants and integrate the remaining exponential.

42:40 Definite Integration by Parts: ∫₁³ (ln x)/x² dx

Rewrite (ln x)/x² as the product ln x · x⁻² before applying parts.
Choose u = ln x and dv = x⁻² dx, giving du = x⁻¹ dx and v = −x⁻¹.
Simplify the remaining product inside the integral using exponent rules, then find the antiderivative.
Apply the bounds 1 and 3 to the entire resulting expression so the final answer contains no x.

48:40 Repeated Integration by Parts for 7x²eˣ

For ∫7x²eˣ dx, choose u = 7x² and dv = eˣ dx; this produces a remaining integral proportional to ∫xeˣ dx.
Apply integration by parts again to ∫xeˣ dx, choosing u = x and dv = eˣ dx.
The result is 7x²eˣ − 14xeˣ + 14eˣ + C; each application lowers the polynomial degree.
The lesson closes by noting that integration by parts and the related reduction techniques are included in the upcoming homework and test.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Prof Staecker.