Choosing the Factors Is the Main Decision
Integration by parts is useful when an integrand is a product, but recognizing a product does not by itself determine how to proceed. The same product can be split in more than one way. A good choice makes the derivative of one factor simpler and leaves a remaining integral that is straightforward to evaluate.
The decision has two parts: choose \(u\), the factor to differentiate, and choose \(dv\), the factor to integrate. It is not enough that \(u\) becomes simpler after differentiation. The antiderivative \(v\) must also be accessible, and the new integral involving \(v\,du\) must be manageable.
The Formula and the Choice Test
For differentiable functions \(u\) and \(v\), the product rule states that \((uv)'=u'v+uv'\). Integrating both sides and rearranging gives the integration-by-parts formula. In the common notation \(du=u'(x)\,dx\) and \(dv=v'(x)\,dx\), it is:
To use this formula, we must be able to differentiate \(u\) and find an antiderivative \(v\) of the factor represented by \(dv\). The resulting integral \(\int v\,du\) must also be one we can evaluate, perhaps after a simple algebraic rewrite.
For a product \(f(x)g(x)\), two natural splits are often possible: \(u=f(x)\), \(dv=g(x)\,dx\), or \(u=g(x)\), \(dv=f(x)\,dx\). Compare them by asking:
- Does differentiating the proposed \(u\) make it simpler?
- Can the proposed \(dv\) be integrated easily?
- Is the new integrand \(v\,du\) easier than the original one?
These questions are more reliable than following a fixed ranking of function types. The best choice depends on the particular product and on the integral that results.
A Derivation That Explains the Rule
The formula is a rearrangement of the product rule, not a separate trick. Here is the derivation with the roles of the factors made explicit.
For differentiable \(u\) and \(v\), \((uv)'=u'v+uv'\).
On an interval where the functions and derivatives involved are continuous, integration gives \(\int (uv)'\,dx=\int u'v\,dx+\int uv'\,dx\).
The left side is \(uv\), up to a constant, so \(uv=\int v\,du+\int u\,dv\), with the constant accounted for in the indefinite integrals.
Subtract \(\int v\,du\) to obtain \(\int u\,dv=uv-\int v\,du\).
This derivation also explains why choosing the factors matters. The rule replaces one product integral with a boundary product \(uv\) and a new integral. A useful choice is one for which that new integral is simpler.
Worked Example: A Logarithm Paired with a Polynomial
Worked Example: Choosing the Logarithm to Differentiate
Evaluate \(\int x\ln x\,dx\) for \(x>0\).
One choice is \(u=\ln x\) and \(dv=x\,dx\). This is promising: \(du=\frac{1}{x}\,dx\), and \(dv\) has the easy antiderivative \(v=\frac{x^2}{2}\). Applying the formula gives:
Check by differentiation:
The derivative of the logarithm removed it from the remaining product, and integrating the polynomial factor was immediate. That combination makes this split effective.
The choice is not useful just because \(\ln x\) has a simple derivative. Its partner \(x\) also has an easy antiderivative, and the new integrand simplifies to a polynomial. All three parts of the choice test are satisfied.
Polynomial Factors Can Decrease in Complexity
A polynomial is often a useful factor to differentiate because each differentiation lowers its degree. When the other factor is a logarithm, differentiating the logarithm can also remove it from the remaining integrand. The next example illustrates both effects, although in this case the logarithm is the factor to differentiate.
Worked Example: A Higher Power Times a Logarithm
Evaluate \(\int x^3\ln x\,dx\) for \(x>0\). Choose \(u=\ln x\), \(dv=x^3\,dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=\frac{x^4}{4}\). Thus:
Differentiate the proposed antiderivative to check it:
Again the remaining integral is a power integral. The original polynomial degree affects the size of \(v\), but does not obstruct the choice because its antiderivative is still simple.
Polynomial factors can also be the factor to differentiate in other products: repeated differentiation can lower their degree. But that is only an advantage if the other factor has an accessible antiderivative and the new integral is manageable. The degree decreasing is not, by itself, enough to guarantee a good choice.
Worked Example: An Inverse-Trigonometric Factor
Worked Example: A Polynomial Times an Inverse-Tangent
Evaluate \(\int x\arctan x\,dx\). Choose \(u=\arctan x\) and \(dv=x\,dx\). The derivative and antiderivative needed for the formula are:
Integration by parts gives:
The remaining rational expression can be simplified by writing \(x^2=(1+x^2)-1\):
Therefore:
Differentiate to verify the result:
The inverse-tangent factor becomes a rational expression when differentiated. Here that expression simplifies to terms with familiar antiderivatives, so the choice works.
In this example, the choice would not be attractive if the derivative of \(\arctan x\) created an integral that was harder than the original. The simplification \(\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}\) is what makes the new integral manageable.
Compare the Whole Outcome, Not One Factor
It can help to write down the anticipated outcome before committing to a split. For the examples above, the comparison looks like this:
| Factor proposed for \(u\) | What happens when it is differentiated? | What must also be checked? |
|---|---|---|
| \(\ln x\) | It becomes \(1/x\), removing the logarithm. | The other factor must have an easy antiderivative. |
| A polynomial | Its degree decreases. | The remaining factor must integrate easily, and the new integral must simplify. |
| \(\arctan x\) | It becomes \(1/(1+x^2)\). | The accompanying factor must lead to a manageable rational expression. |
This comparison is a guide to reasoning, not a rule that automatically identifies the right split. Sometimes both choices work, but one produces fewer steps. Sometimes neither choice makes the remaining integral simpler, in which case another method or an algebraic rewrite may be preferable.
Common Pitfall: Guessing an Antiderivative for the Wrong Factor
A frequent error is to select a factor for \(dv\) without checking its antiderivative. For example, one might try \(u=x\) and \(dv=\ln x\,dx\) in \(\int x\ln x\,dx\), then incorrectly write \(v=\frac{(\ln x)^2}{2}\). That is not an antiderivative of \(\ln x\), because:
The correct antiderivative is \(v=x\ln x-x\), as is checked by differentiating: \(\frac{d}{dx}(x\ln x-x)=\ln x+1-1=\ln x\). The integration-by-parts calculation with this less convenient split is still possible. If \(I=\int x\ln x\,dx\), then:
Rearranging gives \(2I=x^2\ln x-\frac{x^2}{2}\), so \(I=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C\), the result already obtained with the simpler choice. The calculation is valid, but it introduces the original integral again. The more efficient choice avoids that complication.
A Practical Decision Process
When a product suggests integration by parts, use a short comparison rather than choosing automatically:
- List plausible choices for \(u\) and \(dv\).
- For each choice, find \(du\) and \(v\), checking the antiderivative by differentiation if needed.
- Form the new integrand \(v\,du\).
- Choose the split for which the new integral is most manageable, not merely the one with the simplest-looking \(du\).
- Differentiate the final answer to verify that it recovers the original integrand.
This method of choosing parts factors is useful because the formula itself is easy to remember; the reasoning lies in predicting what the transformed integral will look like. Logarithms often simplify when differentiated, polynomial degrees can decrease, and inverse-trigonometric derivatives may produce rational expressions that simplify. Each is a possible signal, subject to checking the other factor and the resulting integral.
Check Your Understanding
For each question, focus on how the factor choice affects the new integral.
- For \(\int x^2\ln x\,dx\), which factor would you choose for \(u\), and what are \(du\) and \(v\)?
- Why is the choice \(dv=\ln x\,dx\) possible in principle but less convenient for \(\int x\ln x\,dx\)?
- In \(\int x\arctan x\,dx\), what identity makes the remaining rational expression easy to integrate?
- What two checks should be made before deciding that a proposed integration-by-parts split is effective?
- Why does differentiating the final answer provide a reliable check on an indefinite integral?
