Integration by Parts using Tabular Approach

Given the integral:

$$ \int x^2 \ln(x) \,dx $$

We’ll use the tabular method to solve it:

1. Choose Functions:
• $$ u = \ln(x) $$
• $$ dv = x^2 \,dx $$
2. Create a Table:

For the table, we will alternate between differentiating \(u\) and integrating \(dv\). Start by listing \(u\), its successive derivatives, and \(dv\), its successive integrals:

$$ u $$	$$ dv $$
$$ \ln(x) $$	$$ x^2 $$
$$ \frac{1}{x} $$	$$ \frac{x^3}{3} $$
$$ -\frac{1}{x^2} $$	$$ \frac{x^4}{12} $$
$$ \frac{2}{x^3} $$	$$ \frac{x^5}{60} $$

3. Apply Signs and Multiply Diagonally:

Now, apply alternating signs down the table. Multiply the terms diagonally and add them up:

$$ (+) \cdot (\ln(x) \cdot x^2) + (-) \cdot \left(\frac{1}{x} \cdot \frac{x^3}{3}\right) + (+) \cdot \left(-\frac{1}{x^2} \cdot \frac{x^4}{12}\right) + (-) \cdot \left(\frac{2}{x^3} \cdot \frac{x^5}{60}\right) $$

Here’s how the multiplication works:

• First row: $$ (+) \cdot (\ln(x) \cdot x^2) = x^2 \ln(x) $$
• Second row: $$ (-) \cdot \left(\frac{1}{x} \cdot \frac{x^3}{3}\right) = -\frac{x^2}{3} $$
• Third row: $$ (+) \cdot \left(-\frac{1}{x^2} \cdot \frac{x^4}{12}\right) = \frac{x^2}{12} $$
• Fourth row: $$ (-) \cdot \left(\frac{2}{x^3} \cdot \frac{x^5}{60}\right) = -\frac{x^2}{30} $$
4. Simplify and Add Up Terms:

$$ \int x^2 \ln(x) \,dx = x^2 \ln(x) – \frac{x^2}{3} – \frac{x^2}{12} – \frac{x^2}{30} + C $$

5. Final Result:

$$ \int x^2 \ln(x) \,dx = x^2 \ln(x) – \frac{13x^2}{60} + C $$