Why the Choice of Factors Matters
Integration by parts is useful when an integrand is a product, but the formula does not prescribe which factor to differentiate and which to integrate. A good choice makes the remaining integral simpler or more familiar. A poor choice can make the calculation longer, or leave an integral that is harder than the one you started with.
The central comparison is between two changes: differentiating a proposed factor \(u\), and finding an antiderivative \(v\) of the proposed factor \(dv\). For example, differentiating a polynomial lowers its degree, while taking an antiderivative of an exponential leaves it exponential. Differentiating a logarithm or inverse trigonometric function often produces an algebraic or rational expression. These tendencies guide the choice, but the resulting integral must still be checked.
The Formula and What a Choice Means
For differentiable functions \(u\) and \(v\), the product rule gives \((uv)'=u'v+uv'\). Rearranging gives \(uv'= (uv)'-u'v\). Integrating both sides on an interval where the functions and integrals are defined gives the indefinite integration by parts formula:
Here \(du=u'(x)\,dx\), and \(dv\) is the part of the integrand whose antiderivative is \(v\). The notation is a way of organizing the product; it does not mean that \(du\) and \(dv\) are independent quantities. To use the formula, the indicated antiderivative \(v\) must exist in a form we can work with, and the remaining integral must be defined on the interval under consideration.
The choice is effective when \(\int v\,du\) is easier than the original integral. It is not enough for \(u\) to become simpler if \(v\) becomes complicated, or if multiplying \(v\) by \(du\) creates a worse integrand.
A Practical Comparison
For products involving the functions in this lesson, some useful tendencies are:
| Factor type | What differentiation tends to do | What antidifferentiation tends to do |
|---|---|---|
| Polynomial | Lowers the degree; a constant eventually differentiates to zero. | Raises the degree by one. |
| Exponential | Keeps the same exponential form, up to a constant factor. | Keeps the same exponential form, up to a constant factor. |
| Logarithm | Often produces a simpler rational expression. | May not have an immediately useful elementary antiderivative in the product at hand. |
| Inverse trigonometric function | Often produces an algebraic expression involving a square root or rational function. | May be inconvenient to integrate directly when it is part of a product. |
This comparison explains several common choices. In a polynomial times exponential product, differentiating the polynomial tends to simplify it, while the exponential is usually straightforward to integrate. In a polynomial times logarithm or inverse trigonometric product, differentiating the logarithm or inverse trigonometric factor often replaces it with a more manageable expression; the polynomial is then integrated. These are starting points, not automatic rules. The final test is the complexity of the remaining integral.
A mnemonic such as LIATE (logarithmic, inverse trigonometric, algebraic, trigonometric, exponential) can help suggest a factor for \(u\). It is only a memory aid. It does not compare the actual resulting integrals, and it does not override the need to check whether \(v\,du\) is simpler.
Worked Example: A Logarithm Paired with an Exponential
Worked Example: A Logarithm Paired with an Exponential
Evaluate \(\int e^x\ln(1+e^x)\,dx\). The logarithmic factor is a good choice for \(u\): its derivative is manageable. The exponential factor is a good choice for \(dv\), since its antiderivative is itself.
Differentiate \(u\) and integrate \(dv\):
Apply integration by parts. Calling the original integral \(I\),
To evaluate the remaining integral, set \(t=1+e^x\). Then \(dt=e^x\,dx\) and \(e^x=t-1\), so
Substituting \(t=1+e^x\) gives \(e^x+1-\ln(1+e^x)\). Therefore, absorbing the constant term into \(C\),
Differentiate the result to check it:
The derivative is the original integrand. The choice was effective because the logarithm differentiated into a manageable factor and the exponential had an immediate antiderivative.
Worked Example: A Polynomial Paired with an Inverse Sine
Worked Example: A Polynomial Paired with an Inverse Sine
Evaluate \(\int x^2\arcsin x\,dx\) for \(-1<x<1\). Choose \(u=\arcsin x\), because its derivative is a standard algebraic expression, and choose \(dv=x^2\,dx\), because it has a simple antiderivative.
Integration by parts gives
For the remaining integral, set \(w=1-x^2\). Then \(dw=-2x\,dx\), and \(x^2=1-w\). Thus
Substitute \(w=1-x^2\) and then return to the integration-by-parts expression:
To verify, differentiate the right-hand side. The derivative of its first term is \(x^2\arcsin x+\frac{x^3}{3\sqrt{1-x^2}}\). The derivatives of the other two nonconstant terms are \(-\frac{x}{3\sqrt{1-x^2}}\) and \(\frac{x}{3}\sqrt{1-x^2}\). Their sum beyond \(x^2\arcsin x\) is
The derivative is \(x^2\arcsin x\), as required. The inverse sine was a good \(u\) because differentiating it exposed a square-root expression that could be handled by substitution.
Worked Example: A Logarithm of a Polynomial
Worked Example: A Logarithm of a Polynomial
Evaluate \(\int x^2\ln(1+x^3)\,dx\) on \(x>-1\), where the logarithm is defined. Take the logarithmic factor as \(u\), and integrate the polynomial factor as \(dv\):
Then
Divide the numerator by the denominator in the remaining rational expression:
Indeed, \(x^2(1+x^3)-x^2=x^5\). Therefore, using \(d(1+x^3)=3x^2\,dx\),
Substitution into the integration-by-parts expression gives
For a check, differentiate. The derivative of the first term is \(x^2\ln(1+x^3)+\frac{x^5}{1+x^3}\); the derivatives of the next two terms are \(-x^2\) and \(\frac{x^2}{1+x^3}\). Hence the derivative is
This choice works because differentiating the logarithm creates a denominator that combines effectively with the integrated polynomial. The polynomial division then reduces the remaining integral to familiar forms.
A Concrete Pitfall: Choosing Factors by Appearance Alone
In the previous example, one might try to set \(u=x^2\) and \(dv=\ln(1+x^3)\,dx\). That choice is not an algebraic error, but it is not useful unless the antiderivative \(v\) can be found conveniently. In particular, the tempting assignment \(v=x\ln(1+x^3)\) is wrong: differentiating it gives
which is not just \(\ln(1+x^3)\). It therefore cannot serve as an antiderivative for the proposed \(dv\). The better choice avoids this problem: integrate \(x^2\) directly and differentiate the logarithm, producing the manageable factor \(\frac{3x^2}{1+x^3}\).
There is a second, more general warning: do not decide only from the factor that looks complicated. Decide by examining the whole proposed next integral \(v\,du\). A factor that differentiates simply may still be a poor \(u\) if the other factor has no useful antiderivative.
A Reliable Selection Process
Write the integrand as a product, and list plausible choices for \(u\) and \(dv\).
Differentiate each candidate for \(u\), and confirm that each candidate for \(dv\) has a usable antiderivative \(v\).
Form \(v\,du\) and ask whether its integral is simpler or more familiar than the original one.
Apply the formula, evaluate the remaining integral, and differentiate the final answer to check it.
Why This Strategy Matters
Integration by parts rearranges a product integral; it does not guarantee a reduction. Choosing the factors well is what makes the method productive. A polynomial often belongs in \(dv\) when paired with a logarithm or inverse trigonometric function, because it is easy to integrate and the other factor often simplifies upon differentiation. With an exponential, antidifferentiation is also straightforward, so the choice depends on whether the other factor becomes simpler and whether the resulting product can be integrated.
When two choices seem plausible, compare the actual remaining integrals rather than relying on a mnemonic. If neither choice produces a manageable next step, integration by parts may not be the right method in that form. When it does apply, the final derivative check catches both an incorrect choice that was not carried through consistently and errors in the calculation.
Check Your Understanding
For each question, explain how differentiation and antidifferentiation affect the proposed factors.
- For a product of a polynomial and an exponential, which factor is often a useful choice for \(u\), and why?
- In \(\int x^2\ln(1+x^3)\,dx\), identify \(u\) and \(dv\), then write \(du\) and \(v\).
- Why is a proposed choice for \(dv\) unsuitable if its antiderivative cannot be found in a useful form?
- What must be examined after choosing \(u\) and \(dv\), before deciding that the choice is effective?
- Why is LIATE a mnemonic rather than a rule that always determines the correct choice?
