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Repeated and Tabular Parts · Tutorial 51 of 998

Tabular Integration by Parts: Layout and Diagonal Products

Build a tabular integration-by-parts layout, track its alternating diagonal products, and handle the remaining integral correctly.

Intermediate 11 min read

What You'll Learn

  • Set up derivative and antiderivative columns for repeated integration by parts
  • Match diagonal products and assign their alternating signs
  • Recognize when a derivative column reaches zero
  • Retain and evaluate the remainder when stopping before the derivative column terminates
  • Check a tabular antiderivative by differentiating it

Why Use a Tabular Layout?

Integration by parts is useful when an integrand is a product and differentiating one factor makes it simpler. In a short problem, the choices and signs are easy to keep track of in a line of working. When the process repeats, however, it is easy to lose a minus sign or pair a derivative with the wrong antiderivative.

A tabular layout organizes that repeated work into two columns. One column records successive derivatives of a chosen factor; the other records successive antiderivatives of the remaining factor. Products formed along matching diagonals are then added with alternating signs. If the derivative column reaches zero, the tabular sum is complete. If it does not, the remaining integral must be included.

Definition: In tabular integration by parts, the derivative column starts with a factor \(u\), and each entry below it is the derivative of the entry above. The antiderivative column starts with \(dv\), and each entry below it is an antiderivative of the entry above. Each diagonal product pairs an entry from the derivative column with the antiderivative entry immediately below and to its right. The signs of successive diagonal products alternate, beginning with \(+\).

The Layout and Its Signs

To build the table, first write the integrand as \(u\,dv\). Choose \(u\) to differentiate and \(dv\) to integrate. Then fill the derivative column downward and the antiderivative column downward. Each antiderivative entry must differentiate to the entry directly above it.

For example, the general shape begins as follows. The first two diagonal products are \(u v_1\) and \(-u'v_2\), where \(v_1\) is an antiderivative of \(dv\), and \(v_2\) is an antiderivative of \(v_1\).

$$ \begin{array}{c|c|c} \text{Sign} & \text{Differentiate} & \text{Integrate} \\ + & u & dv \\ - & u' & v_1 \\ + & u'' & v_2 \\ - & u''' & v_3 \end{array} $$

Read the products diagonally: \(u\) pairs with \(v_1\), \(u'\) pairs with \(v_2\), and \(u''\) pairs with \(v_3\). The sign changes each time: plus, minus, plus, minus. The table is not just a way to list terms. It is a record of repeated applications of the integration-by-parts identity, so the pairing and signs must stay in the same order.

Why the Alternating Signs Appear

The sign pattern begins with the ordinary integration-by-parts formula. If \(v\) is an antiderivative of \(dv\), the product rule gives \((uv)'=u'v+uv'\). Rearranging and integrating yields the identity below. In this form, \(u'\,v\) becomes the new product inside the remaining integral, and the minus sign is the source of the next alternating sign.

$$ \int u\,dv = uv-\int v\,du. $$

This identity also explains why the diagonal is offset: the first antiderivative \(v\) pairs with the original \(u\), while the next antiderivative is paired with \(u'\) after the remaining integral is handled. Each new application introduces one more minus sign. This product-rule argument justifies the first step and the sign change; it does not remove the need to account for any integral left when the table stops.

Important: A tabular sum is a complete antiderivative only when the derivative column terminates at zero, or when the last remaining integral has also been evaluated. Stopping a table does not make its remainder vanish.

Worked Example: A Table That Terminates

Worked Example: Integrating \(x^2\ln x\)

Find an antiderivative of \(x^2\ln x\) for \(x>0\). Choose \(u=x^2\) and \(dv=\ln x\,dx\). The logarithm is integrable, and the polynomial factor eventually differentiates to zero.

First, calculate the derivative column:

$$ x^2,\qquad 2x,\qquad 2,\qquad 0. $$

Now calculate the antiderivative column. The first entry is found by integration by parts, or by differentiating the proposed expression:

$$ v_1=\int \ln x\,dx=x\ln x-x, \qquad v_2=\int (x\ln x-x)\,dx =\frac{x^2}{2}\ln x-\frac{3x^2}{4}. $$

For the next entry, integrate \(v_2\):

$$ v_3=\int\left(\frac{x^2}{2}\ln x-\frac{3x^2}{4}\right)dx =\frac{x^3}{6}\ln x-\frac{x^3}{18}-\frac{x^3}{4} =\frac{x^3}{6}\ln x-\frac{11x^3}{36}. $$

The table pairs entries along the diagonals. Since the derivative column reaches zero, no remainder integral is needed:

$$ \begin{array}{c|c|c} \text{Sign} & \text{Differentiate} & \text{Integrate} \\ + & x^2 & \ln x \\ - & 2x & x\ln x-x \\ + & 2 & \frac{x^2}{2}\ln x-\frac{3x^2}{4} \\ & 0 & \frac{x^3}{6}\ln x-\frac{11x^3}{36} \end{array} $$

Read the diagonal products with their signs:

$$ \begin{aligned} \int x^2\ln x\,dx &=x^2(x\ln x-x) -2x\left(\frac{x^2}{2}\ln x-\frac{3x^2}{4}\right) +2\left(\frac{x^3}{6}\ln x-\frac{11x^3}{36}\right)+C\\ &=\left(1-1+\frac{1}{3}\right)x^3\ln x +\left(-1+\frac{3}{2}-\frac{11}{18}\right)x^3+C\\ &=\frac{x^3}{3}\ln x-\frac{x^3}{9}+C. \end{aligned} $$

Check by differentiating:

$$ \frac{d}{dx}\left(\frac{x^3}{3}\ln x-\frac{x^3}{9}\right) =x^2\ln x+\frac{x^2}{3}-\frac{x^2}{3} =x^2\ln x. $$

The derivative matches the integrand, so the antiderivative is correct for \(x>0\).

Worked Example: Stop and Keep the Remainder

Worked Example: Integrating \(x\ln x\) with One Diagonal

Use \(u=\ln x\) and \(dv=x\,dx\), with \(x>0\). The first derivative of \(u\) is \(1/x\), so this derivative column has not terminated after one step. We will form one diagonal product and explicitly evaluate what remains.

The entries needed are:

$$ \begin{array}{c|c|c} \text{Sign} & \text{Differentiate} & \text{Integrate} \\ + & \ln x & x \\ - & \frac{1}{x} & \frac{x^2}{2} \end{array} $$

The first diagonal contributes \(\frac{x^2}{2}\ln x\). Applying integration by parts once gives the remainder with a minus sign:

$$ \int x\ln x\,dx =\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\cdot\frac{1}{x}\,dx =\frac{x^2}{2}\ln x-\int\frac{x}{2}\,dx. $$

Evaluate the remaining integral:

$$ \int x\ln x\,dx =\frac{x^2}{2}\ln x-\frac{x^2}{4}+C. $$

Check by differentiation:

$$ \frac{d}{dx}\left(\frac{x^2}{2}\ln x-\frac{x^2}{4}\right) =x\ln x+\frac{x}{2}-\frac{x}{2} =x\ln x. $$

The remainder was not zero; it became the integral of \(x/2\), which was then evaluated. Omitting that integral would leave an incomplete answer.

Worked Example: A Second Terminating Table

Worked Example: Integrating \(x\arctan x\)

For \(x\in\mathbb R\), choose \(u=x\) and \(dv=\arctan x\,dx\). The derivative column ends after \(1\), so two diagonal products will complete the result. First compute the antiderivatives required for the right column.

Integration by parts gives the first antiderivative:

$$ v_1=\int\arctan x\,dx =x\arctan x-\frac{1}{2}\ln(1+x^2). $$

Integrate \(v_1\) to obtain the next entry. For its first term, integration by parts gives \(\int x\arctan x\,dx=\frac{x^2+1}{2}\arctan x-\frac{x}{2}\). Also, \(\int\ln(1+x^2)\,dx=x\ln(1+x^2)-2x+2\arctan x\), as can be verified by differentiation. Therefore:

$$ \begin{aligned} v_2 &=\int\left(x\arctan x-\frac{1}{2}\ln(1+x^2)\right)dx\\ &=\frac{x^2+1}{2}\arctan x-\frac{x}{2} -\frac{x}{2}\ln(1+x^2)+x-\arctan x\\ &=\frac{x^2-1}{2}\arctan x+\frac{x}{2} -\frac{x}{2}\ln(1+x^2). \end{aligned} $$

Now assemble the table and its diagonal products:

$$ \begin{array}{c|c|c} \text{Sign} & \text{Differentiate} & \text{Integrate} \\ + & x & \arctan x \\ - & 1 & x\arctan x-\frac{1}{2}\ln(1+x^2) \\ & 0 & \frac{x^2-1}{2}\arctan x+\frac{x}{2}-\frac{x}{2}\ln(1+x^2) \end{array} $$
$$ \begin{aligned} \int x\arctan x\,dx &=x\left(x\arctan x-\frac{1}{2}\ln(1+x^2)\right)-v_2+C\\ &=\frac{x^2+1}{2}\arctan x-\frac{x}{2}+C. \end{aligned} $$

Check by differentiation. Since \(\frac{d}{dx}\arctan x=1/(1+x^2)\),

$$ \begin{aligned} \frac{d}{dx}\left(\frac{x^2+1}{2}\arctan x-\frac{x}{2}\right) &=x\arctan x+\frac{x^2+1}{2(1+x^2)}-\frac{1}{2}\\ &=x\arctan x+\frac{1}{2}-\frac{1}{2} =x\arctan x. \end{aligned} $$

This check confirms both the final expression and the signs in the table.

A Common Mistake: Dropping the Last Integral

A frequent error is to write down the first diagonal product and treat it as the full answer, even though the derivative column has not reached zero. For example, with \(u=\ln x\) and \(dv=x\,dx\), keeping only the first product gives the incorrect claim

$$ \int x\ln x\,dx \overset{\text{wrong}}{=}\frac{x^2}{2}\ln x+C. $$

Differentiating this proposed answer exposes the error:

$$ \frac{d}{dx}\left(\frac{x^2}{2}\ln x\right) =x\ln x+\frac{x}{2}, $$

which is not \(x\ln x\). The missing term must cancel \(x/2\). The table supplies it through the remaining integral, \(-\int x/2\,dx=-x^2/4\). Thus the corrected antiderivative is \(\frac{x^2}{2}\ln x-\frac{x^2}{4}+C\).

Takeaway: A zero at the bottom of the derivative column signals that the table has terminated. If the table is stopped earlier, write the remaining integral with its correct sign and evaluate it separately when possible.

How to Use the Table Reliably

1
Choose the two factors.
Write the integrand as \(u\,dv\). Confirm that the chosen \(dv\) has an antiderivative you can obtain.
2
Fill both columns carefully.
Differentiate downward on the left and integrate downward on the right. Check each entry by differentiation or by taking the derivative of the entry above.
3
Pair along diagonals.
Match each derivative entry with the antiderivative entry one row below it, and assign signs in the order \(+,-,+,-,\ldots\).
4
Check how the table ends.
If the derivative column reaches zero, the diagonal sum is complete. Otherwise, include the remaining integral produced at the stopping point.
5
Verify the result.
Differentiate the proposed antiderivative and confirm that it equals the original integrand on the relevant domain.

Check Your Understanding

For each question, pay attention to the column entries, diagonal pairings, and whether a remainder is present.

  1. If the derivative column begins \(u,u',u''\), what are the first three signs attached to the diagonal products?
  2. In a tabular layout, which antiderivative entry pairs with \(u'\)?
  3. What must be done if the derivative column has not reached zero when you stop forming diagonal products?
  4. For \(\int x\ln x\,dx\), what integral remains after forming the first diagonal product with \(u=\ln x\) and \(dv=x\,dx\)?
  5. Why is differentiating the final expression a useful check on a tabular calculation?

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