The Tabular Method also called the DI Method is an easier way to do integration by parts

It simplifies by making you write less

The way it works is to make a table with 3 columns and a few rows

The first column is S for sign, second column is D for derivatives, and the third is I for integrals

The elements for the sign column are alternating plus and minuses + and -

The first D row is the where your $u$ goes and the the first I is where your $v$ goes

Next you take repeated derivative for $u$ in the D column and repeated anti-derivatives in the I column

For example:

$$\begin{align} \int{(x^3 + 2x)\sin(x) \ dx} \end{align}$$

$$\begin{align} &\begin{array} {c | c | c} S & D & I \\ \hline + & x^3 + 2x & \sin(x)\\ - & 3x^2 + 2 & -\cos(x)\\ + & 6x & -\sin(x)\\ - & 6 & \cos(x)\\ + & 0 & \sin(x) \\ \end{array} \\ \\ \end{align}$$

The final step is to multiply the sign by the derivative, then multiply that by the the integral that is one down

Repeat that until all the way down, adding each term

For example:

$$\begin{align} \int{(x^3 + 2x)\sin(x) \ dx} = -(x^3 + 2x)\cos(x) + (3x^2 + 2)sin(x) + 6xcos(x) + C \end{align}$$

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When Do You Stop?

Stop when the derivative column reaches 0

Stop when the product of the derivative and integral can be integrated easily

Stop when the product of the derivative and integral can be turned into an integral that has been repeated

The reason why we do all 3 of these is because the product of the bottom row is the the integral in the output of integration by parts

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Examples

$$\begin{align} \int{x^3ln(x) \ dx} \end{align}$$

$$\begin{align} &\begin{array} {c | c | c} S & D & I \\ \hline + & ln(x) & x^3 \\ - & 1/x & x^4 / 4\\ \end{array} \\ \\ \end{align}$$

Here we can stop because we can easily the integrate the bottom row

$$\begin{align} \int{x^3ln(x) \ dx} &= \frac{x^4ln(x)}{4} - \int{- \frac{1}{x} \frac{x^4}{4} \ dx} \\ &= \frac{x^4ln(x)}{4} + \int{\frac{x^3}{4} \ dx} \\ &= \frac{x^4ln(x)}{4} + \frac{x^4}{16} + C\\ \end{align}$$

$$\begin{align} \int{e^xsin(3x)\ dx} \end{align}$$

$$\begin{align} &\begin{array} {c | c | c} S & D & I \\ \hline + & e^x & \sin(3x)\\ - & e^x & -\cos(3x)/3\\ + & e^x & -\sin(3x)/9\\ \end{array} \\ \\ \end{align}$$

Here we an stop because we can transform the bottom row into our original integral

$$\begin{align} \int{e^xsin(3x)\ dx} &= -\frac{e^xcos(3x)}{3} + \frac{e^xsin(3x)}{9} + \int{-\frac{e^xsin(3x)}{9} \ dx} \\ \frac{9}{9}\int{e^xsin(3x)\ dx} &= -\frac{e^xcos(3x)}{3} + \frac{e^xsin(3x)}{9} - \frac{1}{9}\int{e^xsin(3x)\ dx} \\ \frac{10}{9}\int{e^xsin(3x)\ dx} &= -\frac{e^xcos(3x)}{3} + \frac{e^xsin(3x)}{9}\\ \int{e^xsin(3x)\ dx} &= \frac{9}{10}\left(-\frac{e^xcos(3x)}{3} + \frac{e^xsin(3x)}{9}\right) + C\\ \end{align}$$

$$\begin{align} \int{x^{12}e^x\ dx} \end{align}$$

$$\begin{align} &\begin{array} {c | c | c} S & D & I \\ \hline + & x^{12} & e^x \\ - & 12x^{11} & e^x\\ + & 132x^{10} & e^x \\ - & 1320x^9 & e^x\\ + & 11880x^8 & e^x \\ - & 95040x^7 & e^x\\ + & 665280x^6 & e^x \\ - & 3991680x^5 & e^x\\ + & 19958400x^4 & e^x \\ - & 79833600x^3 & e^x\\ + & 239500800x^2 & e^x \\ - & 479001600x & e^x\\ + & 479001600 & e^x \\ - & 0 & e^x \\ \end{array} \\ \\ \end{align}$$

Here we stop because the derivative became 0

$$\begin{align} \int{x^{12}e^x\ dx} = x^{12}e^x - 12x^{11}e^x + 132x^{10}e^x - 1320x^9e^x + 11880x^8e^x - 95040x^7e^x + 665280x^6e^x - 3991680x^5e^x \\ +19958400x^4e^x - 79833600x^3e^x + 239500800x^2e^x - 479001600xe^x + 479001600e^x + C \end{align}$$