Tabular Integration

# Tabular Integration

Tabular integration is a special technique for integration by parts that can be applied to certain functions in the form $f(x) = g(x)h(x)$ where one of $g(x)$ or $h(x)$ is can be differentiated multiple times with ease, while the other function can be integrated multiple times with ease. There are two types of Tabular Integration.

The first type is when one of the factors of $f(x)$ when differentiated multiple times goes to $0$.
The second type is when neither of the factors of $f(x)$ when differentiated multiple times goes to $0$.

## Tabular Integration Type 1

Step 1 In the product comprising the function $f$, identify the polynomial and denote it $F(x)$. Denote the other function in the product by $G(x)$. Create a table of $F(x)$ and $G(x)$, and successively differentiate $F(x)$ until you reach $0$. Successively integrate $G(x)$ the same amount of times. Negate every second entry under $F(x)$. Construct the integral by taking the product of $F(x)$ and the first integral of $G(x)$, then add the product of $F'(x)$ times the second integral of $G(x)$, then add the product of $F''(x)$ times the third integral of $G(x)$, etc…

For example, consider the function $f(x) = x^5 \sin x$. Integrating $f$ by integration by parts would be very tedious, so we will use the method of tabular integration. First let $F(x) = x^5$, and let $G(x) = \sin x$. We will now successively differentiate $F(x)$ and integrate $G(x)$ as the following table illustrates:

$F(x)$ $G(x)$
$x^5$ $\sin x$
$5x^4$ $-\cos x$
$20x^3$ $-\sin x$
$60x^2$ $\cos x$
$120x$ $\sin x$
$120$ $-\cos x$
$0$ $-\sin x$

As step 3 states, we will now negate every second entry on the derivatives side of the table:

$F(x)$ $G(x)$
$x^5$ $\sin x$
$-5x^4$ $-\cos x$
$20x^3$ $-\sin x$
$-60x^2$ $\cos x$
$120x$ $\sin x$
$-120$ $-\cos x$
$0$ $-\sin x$

We will now construct our integral by taking the product $F(x)$ and the first antiderivative of $G(x)$, adding it to the product $F'(x)$ by the second antiderivative of $G(x)$, etc…, and therefore:

(1)
\begin{align} \quad \int x^5 \sin x \: dx = -x^5 \cos x + 5x^4\sin x + 20x^3 \cos x - 60x^2 \sin x + -120x \cos x + 120\sin x \\ \quad \int x^5 \sin x \: dx = \cos x (-x^5 + 20x^3 - 120) + \sin x (5x^4 - 60x^2 + 120) + C \end{align}

### Example 1

Evaluate the following integral $\int x^4 e^{-x} \: dx$.

Once again, evaluating this integral would be rather tedious by the process of integration by parts, and thus, the tabular method is much faster. Let $F(x) = x^4$ and $G(x) = e^{-x}$. We now construct our table of derivatives and integrals:

$F(x)$ $G(x)$
$x^4$ $e^{-x}$
$4x^3$ $-e^{-x}$
$12x^2$ $e^{-x}$
$24x$ $-e^{-x}$
$24$ $e^{-x}$
$0$ $-e^{-x}$

We will now negate every second entry under $F(x)$ to get:

$F(x)$ $G(x)$
$x^4$ $e^{-x}$
$-4x^3$ $-e^{-x}$
$12x^2$ $e^{-x}$
$-24x$ $-e^{-x}$
$24$ $e^{-x}$
$0$ $-e^{-x}$

And we construct our integral as:

(2)
\begin{align} \int x^4 e^{-x} \: dx = -x^4e^{-x} - 4x^3e^{-x} - 12x^2e^{-x} -24xe^{-x} - 24e^{-x} \\ \int x^4 e^{-x} \: dx = -e^{-x}(x^4 + 4x^3 + 12x^2 + 24x + 24) + C \end{align}