Why the Choice of Factors Matters
Integration by parts is especially effective when an integrand is the product of a polynomial and an exponential function. The key is to assign the polynomial to \(u\) and the exponential factor, including \(dx\), to \(dv\). Differentiating a polynomial lowers its degree, while integrating an exponential preserves its basic form. Repeating the process therefore simplifies the remaining integral.
For example, if the polynomial is quadratic, its first derivative is linear, its second derivative is constant, and its third derivative is zero. This gives a natural stopping point: after enough rounds of integration by parts, the polynomial part disappears.
The Formula and the Factor Choice
The integration-by-parts formula follows from the product rule and is used here in its indefinite-integral form:
In this formula, \(u\) and \(v\) are functions of \(x\), and \(du\) and \(dv\) represent their differentials. When using the formula for an indefinite integral, include a constant of integration in the final result.
For \(e^{ax}\), where \(a\ne 0\), an antiderivative is \(\frac{1}{a}e^{ax}\), since
Thus, for an integrand \(P(x)e^{ax}\), where \(P\) is a polynomial, the factor choice is typically
This choice is not a rule that must be followed mechanically in every possible integral. It is effective here because the derivative of the polynomial becomes simpler, while the exponential is straightforward to integrate. Choosing the exponential as \(u\) and the polynomial as \(dv\) generally makes the remaining integral more complicated rather than reducing the polynomial degree.
Worked Example: A Linear Polynomial and an Exponential
Worked Example: A Linear Polynomial and an Exponential
Evaluate \(\int x e^{2x}\,dx\).
Choose \(u=x\), because its derivative is the simpler function \(1\). Choose \(dv=e^{2x}\,dx\), whose antiderivative is \(\frac12e^{2x}\). Therefore,
Apply integration by parts:
Check by differentiating the proposed antiderivative:
The derivative is the original integrand, so the antiderivative checks.
Why Repeating the Choice Works
After one application of integration by parts, the polynomial is differentiated once. If it is still nonconstant, the same choice can be used again on the remaining polynomial-exponential product. Each new derivative reduces the polynomial's degree by one, so after finitely many steps the polynomial derivative becomes zero.
The exponential remains an exponential throughout: each integration of \(e^{ax}\) contributes a factor of \(\frac1a\), and each differentiation of the exponential contributes a factor of \(a\). The method therefore transforms the integral into a finite sequence of simpler polynomial-exponential terms rather than creating an endless cycle.
Worked Example: A Quadratic Polynomial
Worked Example: A Quadratic Polynomial
Evaluate \(\int x^2e^x\,dx\).
Set \(u=x^2\) and \(dv=e^x\,dx\). Then \(du=2x\,dx\) and \(v=e^x\). The first application gives
The remaining integral still has a polynomial times an exponential, so apply integration by parts again. For \(\int xe^x\,dx\), choose \(u=x\), \(du=dx\), \(dv=e^x\,dx\), and \(v=e^x\):
Substitute this result into the first equation:
Differentiate to verify the result:
The terms of degree one and zero cancel in the derivative, leaving the original quadratic-exponential integrand.
Worked Example: A Polynomial with a Negative Exponential Rate
Worked Example: A Polynomial with a Negative Exponential Rate
Evaluate \(\int (3x^2-2x+1)e^{-2x}\,dx\). The same factor choice applies: take the polynomial as \(u\) and \(e^{-2x}\,dx\) as \(dv\). The negative rate matters because \(\int e^{-2x}\,dx=-\frac12e^{-2x}\).
For the first application, use
Then
For the remaining integral, let \(u=3x-1\), \(du=3\,dx\), \(dv=e^{-2x}\,dx\), and \(v=-\frac12e^{-2x}\). Thus
Substituting and collecting terms gives
Check by differentiating. Let \(Q(x)=-\frac32x^2-\frac12x-\frac34\), so \(Q'(x)=-3x-\frac12\). The product rule gives
This is exactly the original integrand, including its exponential factor.
Choosing Factors: A Practical Comparison
The choice of \(u\) and \(dv\) determines whether the remaining integral is simpler. For polynomial-exponential products, compare what happens to the polynomial under each possible choice.
| Choice | Effect after integration by parts | Typical usefulness here |
|---|---|---|
| \(u=P(x)\), \(dv=e^{ax}\,dx\) | \(du=P'(x)\,dx\), so the polynomial degree decreases | Usually effective when \(a\ne 0\) |
| \(u=e^{ax}\), \(dv=P(x)\,dx\) | \(du=ae^{ax}\,dx\), while \(v\) is an antiderivative of the polynomial | Usually less effective because the polynomial's degree can increase |
The second choice is not automatically invalid, but it often works against the aim of simplifying the integral. For instance, integrating \(x\) to obtain \(\frac{x^2}{2}\) increases the polynomial degree, whereas differentiating \(x\) gives the simpler constant \(1\).
A Common Sign Error
A frequent mistake is to replace the minus sign in integration by parts with a plus sign. For \(\int xe^x\,dx\), the incorrect calculation would be
This answer fails a differentiation check:
which is not the original integrand \(xe^x\). The correct formula is \(uv-\int v\,du\), so the corrected calculation is
Now the derivative check confirms it:
What to Check Before Finishing
For an indefinite integral of a polynomial times \(e^{ax}\), use the polynomial for \(u\) and integrate the exponential for \(dv\), provided \(a\ne 0\). If the polynomial still has positive degree after an application, repeat the process on the remaining integral. Keep the factor \(\frac1a\) every time you integrate \(e^{ax}\), and preserve the minus sign in the integration-by-parts formula.
Finally, differentiate the proposed antiderivative. This is particularly useful when several terms have been collected: the product rule exposes errors in coefficients, signs, or the exponential rate. The derivative must equal the entire original integrand, not merely its polynomial or exponential factor.
Check Your Understanding
For each question, state the factor choice clearly and use differentiation when asked to verify an antiderivative.
- For \(\int x^3e^x\,dx\), what should \(u\) and \(dv\) be, and how many derivatives of the polynomial are needed before it becomes zero?
- Evaluate \(\int xe^{3x}\,dx\) by integration by parts, then differentiate your result.
- In \(\int (x^2+1)e^{-x}\,dx\), what is \(v\) when \(dv=e^{-x}\,dx\)?
- Explain why choosing the polynomial as \(u\) makes the repeated process terminate.
- For \(\int xe^x\,dx\), show what derivative check reveals about the incorrect answer \(xe^x+e^x+C\).
