Integration by Parts Notes

Overview of Integration by Parts

• Core formula: $\int u\,dv = uv - \int v\,du.$
• The goal is to pick u and dv so that:
  • du becomes simpler (easier to integrate when differentiated), and
  • v is easy to multiply by du and integrate in the remaining integral.

Choosing u and dv: Practical guidelines from the lecture

• Rule of thumb emphasized:
  • Do not choose dv to involve a term like $x^2$ if, after integrating, the resulting $v$ would introduce higher powers of $x$ that complicate the remaining integral.
  • Specifically, if you take $u$ as a simple function and $dv$ as $x^2\,dx$, then $v = \frac{x^3}{3}$, which can lead to a remaining integral with larger powers when multiplied by $du$.
  • The speaker suggests avoiding setting $\text{(part)}\,dv$ to $x^2$ for this kind of format, because it tends to escalate complexity rather than reduce it.
• Implication: choose $u$ so that differentiating it reduces the complexity (degree, form), and choose $dv$ so that its integral $v$ does not blow up powers of $x$ in the residual integral.

About powers of x and negative exponents

• When differentiating a power of $x$:
  • If $u = x^m$, then $\frac{du}{dx} = m\,x^{m-1}.$
• If the exponent is negative, or the term is a reciprocal like $x^{-1} = \frac{1}{x}$:
  • Differentiating increases the negative magnitude: for example, if $x^p$ with $p$ negative, then $\frac{d}{dx}x^p = p\,x^{p-1}.$ This tends to produce even more negative exponents.
• Important note on $x^{-1}$ (i.e., $\frac{1}{x}$):
  • The integral is not a simple power; rather, $\int x^{-1} \,dx = \ln|x| + C.$
  • When you see $x^{-1}$ in the integrand, you may encounter a logarithmic antiderivative for its integral.
• Example from the lecture language:
  • For a term like $x^{-5}$, differentiating $x^{-6}$ would yield a multiple of $x^{-6}$; the pattern shows negative exponents can get unwieldy under repeated differentiation, which reinforces the need to choose $u$ and $dv$ thoughtfully when negative powers are present.
• Summary rule to remember: if you see a negative exponent or a reciprocal form, recognize that differentiation will push the exponent downward and be mindful of how this interacts with the remaining integral after applying the parts formula.

The integration-by-parts formula in use

• General formula reminder: $\int u\,dv = uv - \int v\,du.$
• In the example discussion, the segment "we subtract the second part of the integral" refers to the remaining integral: the term $\int v\,du.$
• The signs and constants come from the chosen $u$ and $dv$; in the specific example, the instructor notes:
  • $v_x$ turns out to be a negative coefficient (e.g., $-\frac{1}{4}$) in the worked setup,
  • $u'_x$ (i.e., $\frac{du}{dx}$) turns out to be a positive coefficient (e.g., $\frac{1}{4}$),
  • but the exact numbers depend on the particular choice of $u$ and $dv$ in the problem.
• Takeaway: the structure to memorize is $\int u\,dv = uv - \int v\,du.$ The numeric coefficients arise from your chosen $u$ and $dv$.

Worked illustrative scenarios

• Scenario A: ∫ x^2 cos x dx
  • If you take $u = x^2$ and $dv = \cos x\,dx$, then
  • $du = 2x\,dx$, $v = \sin x$.
  • Result: $\int x^2 \cos x\,dx = x^2 \sin x - \int 2x \sin x\,dx.$
  • The remaining integral is easier (still needs another IBP), but the key is you reduced the power of $x$ in the remaining integrand after each step.
  • Why the caution about $dv = x^2 dx$? If you instead tried $u = $something that makes $du$ simple but $dv = x^2 dx$ gives a very large power in $v$, the leftover integral can become unwieldy. The lesson is to balance the reduction in $du$ with a manageable $v$.
• Scenario B: ∫ (ln x)/x dx
  • Take $u = \ln x$, $dv = x^{-1} dx$.
  • Then $du = \frac{1}{x} dx$, $v = \ln x$.
  • Apply IBP: $\int \frac{\ln x}{x}\,dx = (\ln x)(\ln x) - \int \ln x \cdot \frac{1}{x} dx = \ln^2 x - \int \frac{\ln x}{x} dx.$
  • Solve for the integral: $\int \frac{\ln x}{x} dx = \frac{1}{2}\ln^2 x + C.$
  • This example shows how choosing $dv = x^{-1}$ leads to a logarithmic antiderivative and how IBP can yield a solvable equation for the integral.

Connections to foundational principles

• Power rule and generalized power behavior:
  • Differentiation reduces the exponent by 1: $\frac{d}{dx}x^m = m x^{m-1}$.
  • Integration increases the exponent by 1 (except for the case $m=-1$ which yields a log).
• Reciprocal form and logs:
  • When encountering $x^{-1}$, be prepared to obtain a logarithmic antiderivative: $\int x^{-1}\,dx = \ln|x| + C.$
• Strategy alignment with calculus basics:
  • IBP is a tool to transfer differentiation from a hard part to a part that is easier to integrate, while maintaining a balance to avoid escalating complexity.

Quick reference reminders (formulae in LaTeX)

• Integration by parts: $\int u\, dv = uv - \int v\, du.$
• Derivative of a power: $\frac{d}{dx} x^m = m\,x^{m-1}.$
• Integral of a negative power (excluding $m=-1$): $\int x^m\,dx = \frac{x^{m+1}}{m+1} + C, \quad m \neq -1.$
• Integral when the power is $-1$: $\int x^{-1}\,dx = \ln|x| + C.$
• Example result pattern for combined powers (illustrative): for $u = x^m$, $dv = x^n dx$ with $m,n \neq -1$,
  • $\int u\,dv = \frac{x^{m+n+1}}{n+1} - \frac{m}{(n+1)(m+n+1)} x^{m+n+1} + C.$
  • Note: This is one way to summarize the effect after applying IBP; the exact intermediate steps depend on your chosen $u$ and $dv$.

Practical tips for exam prep

• Before starting IBP, quickly assess whether differentiating $u$ will simplify the expression and whether integrating $dv$ will yield a straightforward antiderivative.
• Be mindful of negative exponents; expect potential logarithmic forms when integrating $x^{-1}$.
• Practice with a mix of polynomial, exponential, and trigonometric parts to reinforce the decision-making process for $u$ and $dv$.
• Always write the IBP steps clearly: identify $u$, compute $du$, identify $dv$, compute $v$, then form and simplify $uv - \int v\,du$.