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Repeated and Tabular Parts · Tutorial 52 of 998

Deriving the Tabular Formula from Repeated Integration by Parts

Learn how repeated integration by parts produces the tabular formula, and how to identify and check its remainder.

Intermediate 10 min read

What You'll Learn

  • State the hypotheses for a finite repeated integration-by-parts formula
  • Derive the alternating diagonal products from successive applications of the parts identity
  • Write the remainder after any chosen number of integrations
  • Identify when the remainder vanishes because the differentiated factor becomes zero
  • Check a tabular result by differentiating it
  • Recognize why an unfinished table does not justify dropping its remainder

Why Derive the Tabular Formula?

Tabular integration by parts organizes repeated applications of the integration-by-parts identity into two columns: successive derivatives of one factor and successive antiderivatives of the other. The products along matching rows are called diagonal products. Their alternating signs are not a separate rule to memorize; they follow from the minus sign in integration by parts each time the remaining integral is treated again.

The formula also has a remainder. A table gives an antiderivative directly when that remainder vanishes, as it does when the differentiated factor eventually becomes zero. Otherwise, the formula is still an identity, but the remainder must be retained or evaluated by another method.

Central idea: After each integration by parts, the new integral contains one more derivative of the first factor and one more antiderivative of the second. This produces the next diagonal product and reverses its sign.

Setting Up the Notation

Let \(I\) be an interval. Suppose \(u\) has continuous derivatives through order \(r\) on \(I\), where \(r\) is a positive integer. Let \(g\) be continuous, and choose functions \(V_0,V_1,\ldots,V_{r-1}\) on \(I\) such that

$$ V_0'=g \qquad\text{and}\qquad V_{k+1}'=V_k \quad (0\le k\le r-2). $$

Thus \(V_0\) is an antiderivative of the original second factor \(g\), \(V_1\) is an antiderivative of \(V_0\), and the chain continues. Write \(u^{(k)}\) for the \(k\)-th derivative of \(u\), with \(u^{(0)}=u\).

Definition: The finite repeated integration-by-parts formula after \(r\) applications is
$$ \int u(x)g(x)\,dx = \sum_{k=0}^{r-1}(-1)^k u^{(k)}(x)V_k(x) + (-1)^r\int u^{(r)}(x)V_{r-1}(x)\,dx. $$
The last integral is the remainder. As with any indefinite-integral identity, the equality is understood up to an additive constant.

The formula can be used on any interval where the stated functions and derivatives exist. In particular, if a factor has a singularity, work on an interval that does not cross that singularity.

Why the Diagonal Products Appear

Start with the usual integration-by-parts identity, \(\int a\,db=ab-\int b\,da\). In this setting, \(V_0'=g\), so the first application gives

$$ \int ug\,dx = uV_0-\int u'V_0\,dx. $$

The remaining integral has a minus sign in front of it. Since \(V_1'=V_0\), applying integration by parts to that remaining integral gives

$$ \int u'V_0\,dx = u'V_1-\int u''V_1\,dx. $$

Substituting this into the first line changes the sign of the new product and leaves a remainder with a plus sign:

$$ \int ug\,dx = uV_0-u'V_1+\int u''V_1\,dx. $$

The next application produces \(-u''V_2\), because the remainder just obtained has a plus sign and integration by parts itself subtracts the next integral. Each iteration therefore adds one derivative, one antiderivative, and one sign change. The products \(u^{(k)}V_k\) match the derivative and antiderivative in the same row of a tabular layout, which is why they are the diagonal products.

Derivation of the Finite Formula

The pattern can be justified by induction on the number \(r\) of applications. For one application, the integration-by-parts identity gives \(uV_0-\int u'V_0\,dx\), which is the stated formula with \(r=1\).

Now suppose the formula holds after \(r\) applications. Its remainder is \((-1)^r\int u^{(r)}V_{r-1}\,dx\). Since \(V_r'=V_{r-1}\), one more application of integration by parts gives

$$ \int u^{(r)}V_{r-1}\,dx = u^{(r)}V_r-\int u^{(r+1)}V_r\,dx. $$

Multiplying by \((-1)^r\), the new product is \((-1)^r u^{(r)}V_r\), and the new remainder is \((-1)^{r+1}\int u^{(r+1)}V_r\,dx\). These are exactly the next term and remainder in the formula with \(r+1\) applications. This proves the formula for every positive integer \(r\) for which the required functions exist.

1
Differentiate the first factor.
The successive entries are \(u,u',u'',\ldots\).
2
Integrate the second factor repeatedly.
The entries begin with \(V_0\), where \(V_0'=g\), and continue with \(V_{k+1}'=V_k\).
3
Pair matching rows and alternate signs.
The \(k\)-th product is \((-1)^k u^{(k)}V_k\).
4
Keep the remainder unless it is zero or has been evaluated.
After \(r\) applications it is \((-1)^r\int u^{(r)}V_{r-1}\,dx\).

When Does the Remainder Vanish?

The remainder after \(r\) applications is zero if \(u^{(r)}(x)V_{r-1}(x)=0\) throughout the interval. The most common useful case is \(u^{(r)}=0\) throughout that interval. For example, if \(u\) is a polynomial of degree less than \(r\), its \(r\)-th derivative is zero. The diagonal products then give the complete antiderivative, up to a constant.

A remainder is not zero merely because a table has reached a convenient number of rows. If \(u^{(r)}\) is not zero, the last integral is still part of the identity. It may sometimes be evaluated separately, or further integration by parts may help, but neither possibility permits dropping it without justification.

Worked Examples

Worked Example: A Table That Terminates

Find \(\int \frac{x^2}{(x+1)^2}\,dx\), working on an interval that does not contain \(x=-1\). Choose \(u=x^2\) and \(g=(x+1)^{-2}\). The derivative column is

$$ u=x^2,\qquad u'=2x,\qquad u''=2,\qquad u'''=0. $$

For the other column, integrate successively:

$$ V_0=-\frac{1}{x+1},\qquad V_1=-\ln|x+1|,\qquad V_2=-(x+1)\ln|x+1|+(x+1). $$

Indeed, \(V_0'=(x+1)^{-2}\), \(V_1'=V_0\), and \(V_2'=V_1\). Since \(u'''=0\), three applications leave no remainder. The diagonal products give

$$ \begin{aligned} \int \frac{x^2}{(x+1)^2}\,dx &=uV_0-u'V_1+u''V_2\\ &=-\frac{x^2}{x+1}+2x\ln|x+1|\\ &\quad+2\bigl(-(x+1)\ln|x+1|+(x+1)\bigr)\\ &=-\frac{x^2}{x+1}-2\ln|x+1|+2x+2\\ &=x+3-\frac{1}{x+1}-2\ln|x+1|+C. \end{aligned} $$

To check, differentiate the simplified result:

$$ \begin{aligned} \frac{d}{dx}\left(x+3-\frac{1}{x+1}-2\ln|x+1|\right) &=1+\frac{1}{(x+1)^2}-\frac{2}{x+1}\\ &=\frac{(x+1)^2+1-2(x+1)}{(x+1)^2}\\ &=\frac{x^2}{(x+1)^2}. \end{aligned} $$

The derivative agrees with the original integrand on either interval where \(x\ne -1\).

Worked Example: Two Diagonals and a Vanishing Derivative

Find \(\int \frac{x}{x+1}\,dx\), with \(x\ne -1\). Take \(u=x\) and \(g=(x+1)^{-1}\). Then \(u'=1\) and \(u''=0\). Choose \(V_0=\ln|x+1|\) and \(V_1=(x+1)\ln|x+1|-(x+1)\), so \(V_0'=g\) and \(V_1'=V_0\). The formula after two applications is

$$ \begin{aligned} \int \frac{x}{x+1}\,dx &=uV_0-u'V_1+\int u''V_1\,dx\\ &=x\ln|x+1|-\bigl((x+1)\ln|x+1|-(x+1)\bigr)\\ &=x+1-\ln|x+1|+C. \end{aligned} $$

The remainder is zero because \(u''=0\). Check by differentiation:

$$ \frac{d}{dx}\bigl(x+1-\ln|x+1|\bigr) =1-\frac{1}{x+1} =\frac{x}{x+1}. $$

This check confirms both the result and the minus sign between the two diagonal products.

Worked Example: Retaining a Nonzero Remainder

Apply two integrations by parts to \(\int \frac{1}{x+1}\,dx\), on an interval not containing \(-1\). Set \(u=(x+1)^{-1}\) and \(g=1\). Then

$$ u'=-\frac{1}{(x+1)^2},\qquad u''=\frac{2}{(x+1)^3},\qquad V_0=x,\qquad V_1=\frac{x^2}{2}. $$

Here \(V_0'=1\) and \(V_1'=V_0\). The two-application formula gives

$$ \begin{aligned} \int \frac{1}{x+1}\,dx &=uV_0-u'V_1+\int u''V_1\,dx\\ &=\frac{x}{x+1}+\frac{x^2}{2(x+1)^2} +\int \frac{x^2}{(x+1)^3}\,dx. \end{aligned} $$

In this example \(u''\) is not zero, so the remainder cannot be discarded. To check the identity, differentiate its right-hand side. The first two terms have derivatives \(1/(x+1)^2\) and \(x/(x+1)^3\), respectively. The derivative of the remaining integral is \(x^2/(x+1)^3\). Their sum is

$$ \frac{1}{(x+1)^2}+\frac{x}{(x+1)^3}+\frac{x^2}{(x+1)^3} = \frac{x+1+x+x^2}{(x+1)^3} = \frac{1}{x+1}. $$

Thus the repeated-parts identity is verified. It has not, by itself, evaluated the remaining integral; it has expressed the original integral in terms of a remainder that must still be handled.

A Common Sign Error

In \(\int \frac{x}{x+1}\,dx\), a common mistake is to add both diagonal products, writing \(xV_0+V_1\). That gives

$$ x\ln|x+1|+(x+1)\ln|x+1|-(x+1) =(2x+1)\ln|x+1|-(x+1). $$

This is not an antiderivative. Its derivative is

$$ 2\ln|x+1|+\frac{2x+1}{x+1}-1, $$

which is not \(x/(x+1)\). The correction is to alternate signs: the first product has a plus sign, and the second has a minus sign because it comes from the subtracted integral in the first integration-by-parts step.

Using the Formula Reliably

Tabular integration is most useful when repeated differentiation simplifies or eventually eliminates the chosen first factor. Before stopping, check the derivative column: if its last entry is not zero, write down the remainder explicitly. Also check the antiderivative column by differentiating each entry back to the one above it. These checks catch both sign errors and mistakes in the repeated antiderivatives.

Takeaway: A tabular formula is repeated integration by parts written compactly. Its diagonal products and alternating signs come from iteration, and its remainder records exactly what has not yet been integrated.

Check Your Understanding

Use the formula and its conditions to answer these questions.

  1. After \(r\) applications of integration by parts, what is the sign and form of the remainder?
  2. Why does each successive diagonal product have the opposite sign from the preceding one?
  3. What condition on \(u\) guarantees that the remainder vanishes after \(r\) applications?
  4. If \(u^{(r)}\) is not zero, is it valid to omit the remainder? Explain.
  5. For \(\int \frac{x}{x+1}\,dx\), what does differentiating the incorrect expression \(xV_0+V_1\) reveal?

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