Ma · Mathematics

How to Use the Tabular Method for Integration by Parts

Published By the 9to5Science editorial team

At a glance

  1. 1. Choose the factors

    Write the integrand as a product, choosing the factor to differentiate and the factor to integrate.

  2. 2. Build the columns

    Differentiate the first factor repeatedly and integrate the second factor repeatedly.

  3. 3. Add diagonal products

    Multiply diagonally, assign alternating signs, and add the resulting terms.

  4. 4. Check the remainder

    Stop when the derivatives reach zero; otherwise include the remaining integral.

Title: How to Use the Tabular Method for Integration by Parts Category: mathematics

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The tabular method is a compact way to apply integration by parts repeatedly. It is most useful when one factor becomes simpler each time you differentiate it—especially when that factor is a polynomial—and the other factor has easy antiderivatives.

The method does not change the integration-by-parts rule. It simply organizes the repeated steps so you can see the signs and products at a glance.

The idea behind the table

Integration by parts says

∫u dv=uv−∫v du.\int u\,dv=uv-\int v\,du.

For a product f(x)g(x)f(x)g(x), choose ff as the factor to differentiate and gg as the factor to integrate. The first application produces a product f∫g dxf\int g\,dx and a new integral involving f′f'. Applying the rule again to that new integral introduces f′′f'', and so on. The tabular layout records these derivatives, antiderivatives, products, and alternating signs in one place.

A useful rule of thumb is to choose the factor that becomes simpler under differentiation as the one to differentiate. The other factor should be straightforward to integrate repeatedly. For example, a polynomial paired with an exponential or sine function is often a good fit. Heuristics such as LIATE (logarithmic, inverse trigonometric, algebraic, trigonometric, exponential) can help choose which factor to differentiate, but they are not a law; check that the resulting antiderivatives remain manageable.

How to make the table

For an integral of the form ∫f(x)g(x) dx\int f(x)g(x)\,dx:

  1. Write successive derivatives of ff in the left column, starting with ff.
  2. Write successive antiderivatives of gg in the right column, starting with an antiderivative of gg.
  3. Pair corresponding entries: ff with the first antiderivative, f′f' with the second, and so on. In common tabular layouts, these pairings are shown diagonally.
  4. Give the products alternating signs, starting with plus, and add them.

The first product is positive because the first integration-by-parts step gives f∫g dxf\int g\,dx. The next product is negative, the next positive, and so forth. If a derivative becomes zero, the process ends and there is no leftover integral. If it does not, the unused part of the table represents a remaining integral, with the appropriate alternating sign.

Worked example: integrate x2exx^2e^x

Evaluate

∫x2ex dx.\int x^2e^x\,dx.

Differentiate x2x^2 repeatedly. Integrate exe^x repeatedly; every antiderivative is exe^x.

SignDerivatives of x2x^2Antiderivatives of exe^x
++x2x^2exe^x
−-2x2xexe^x
++22exe^x
00

Multiply the corresponding entries, starting with the first row:

∫x2ex dx=x2ex−(2x)ex+(2)ex+C=ex(x2−2x+2)+C.\begin{aligned} \int x^2e^x\,dx &=x^2e^x-(2x)e^x+(2)e^x+C\\ &=e^x(x^2-2x+2)+C. \end{aligned}

The derivative column reaches zero, so there is no remaining integral. To check the result, differentiate ex(x2−2x+2)e^x(x^2-2x+2). The product rule gives ex(x2−2x+2)+ex(2x−2)=x2exe^x(x^2-2x+2)+e^x(2x-2)=x^2e^x, as required.

When the table does not end

If repeated differentiation does not reach zero, the method may still help, but you must keep the remaining integral. For example, differentiating sin⁡x\sin x continues in a cycle rather than reaching zero. Tabular integration can organize that cycle, but it does not automatically eliminate the integral. In cases such as ∫exsin⁡x dx\int e^x\sin x\,dx, repeated integration by parts returns the original integral; you then solve for it algebraically.

For a definite integral, use the same diagonal products and evaluate their sum at the limits. If the table ends, then

∫abf(x)g(x) dx=[∑k=0m(−1)kf(k)(x)Gk+1(x)]ab,\int_a^b f(x)g(x)\,dx =\left[\sum_{k=0}^{m}(-1)^k f^{(k)}(x)G_{k+1}(x)\right]_a^b,

where G1G_1 is an antiderivative of gg, each later GG is an antiderivative of the one before it, and f(m+1)=0f^{(m+1)}=0. The brackets mean evaluate the entire expression at bb and subtract its value at aa.

Common mistakes

  • Pairing entries incorrectly. In the displayed layout, pair each derivative with the antiderivative in the same row: ff with G1G_1, f′f' with G2G_2, and so on. In a layout that shifts the columns, these pairings may be shown diagonally.
  • Forgetting the alternating signs. The signs start with plus and then switch: +,−,+,−,…+,-,+,-,\ldots.
  • Stopping too early. You can omit the remaining integral only if the derivative column reaches zero. Otherwise, write that integral with its sign.
  • Choosing a difficult factor to integrate. Repeated antiderivatives must also stay manageable. A factor that looks convenient at first may make the table less useful if its antiderivatives become complicated.
  • Omitting the constant. For an indefinite integral, add CC after the final expression.

References

  1. OpenStax, Calculus Volume 2, Section 3.1: Integration by Parts

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