AP Calculus BC

Integral of x^2 ln x: Why ln x Should Be u

Q: What is the integral of x^2 ln x?

It is \frac{x^{3}\ln x}{3} - \frac{x^{3}}{9} + C, valid for x > 0 since \ln x is undefined otherwise.

Q: Why is u = ln x and not x^2?

Because \ln x has no antiderivative you would want as v, and differentiating it gives \frac{1}{x}, which reduces the leftover integral to \int x^{2}\,dx.

The integral of x squared times ln x is x cubed ln x over 3, minus x cubed over 9, plus C. Integration by parts is the technique, and ln x is the choice that finishes in one pass: it differentiates to 1/x, which cancels most of the leftover polynomial.

\int x^{2}\ln x\,dx = \frac{x^{3}\ln x}{3} - \frac{x^{3}}{9} + C

Choosing u and dv

Integration by parts needs the product $x^{2}\ln x$ split into $u$ and $dv$ . LIATE ranks the logarithm ahead of the algebraic factor, so $u = \ln x$ and $dv = x^{2}\,dx$ .

That gives $du = \frac{1}{x}\,dx$ and $v = \frac{x^{3}}{3}$ .

\int x^{2}\ln x\,dx = \frac{x^{3}\ln x}{3} - \int \frac{x^{3}}{3}\cdot\frac{1}{x}\,dx

= \frac{x^{3}\ln x}{3} - \frac{1}{3}\int x^{2}\,dx = \frac{x^{3}\ln x}{3} - \frac{x^{3}}{9} + C

The leftover integral is a plain power, which is the whole reason for that choice. Differentiating $\ln x$ removed the logarithm from the problem, and the $\frac{1}{x}$ it produced cancelled a power off $x^{3}$ .

Compare it with x^2 e^x

In $\int x^{2}e^{x}\,dx$ the polynomial is $u$ , because differentiating $x^{2}$ grinds it down while $e^{x}$ integrates back to itself. The ranking flips here for one reason: $\ln x$ is the factor you do not want to integrate, and it is the factor that gets simpler when differentiated.

So the rule underneath both problems is the same. Take as $u$ whichever factor improves under differentiation, and as $dv$ whichever factor you can antidifferentiate on sight. Whenever $\ln x$ sits in a product, it wins the $u$ slot.

Count the passes

x^2 ln x needs parts once. x^2 e^x needs parts twice, because the polynomial has to be differentiated down to a constant. If your choice of u leads to a second pass here, you chose the slower u.

The mistakes students make

Two of these three give answers that differentiate to something visibly close to $x^{2}\ln x$ , which is why they survive a quick glance.

Setting $u = x^{2}$ and $dv = \ln x\,dx$ . That needs $\int\ln x\,dx = x\ln x - x$ before the first line can be written, and the leftover integral $\int 2x\left(x\ln x - x\right)dx$ reproduces the original integral, so you have to solve a circular equation instead of finishing in one line.
Dropping the $\frac{1}{x}$ from $du$ , which turns the second integral into $\int\frac{x^{3}}{3}\,dx$ and produces $\frac{x^{3}\ln x}{3} - \frac{x^{4}}{12}$ .
Stopping at $\frac{x^{3}\ln x}{3} - \frac{x^{3}}{3}$ . Integrating $x^{2}$ contributes a second factor of $\frac{1}{3}$ , so the denominator of the last term is $9$ .

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of x^2 ln x?

It is $\frac{x^{3}\ln x}{3} - \frac{x^{3}}{9} + C$ , valid for $x > 0$ since $\ln x$ is undefined otherwise.

Why is u = ln x and not x^2?

Because $\ln x$ has no antiderivative you would want as $v$ , and differentiating it gives $\frac{1}{x}$ , which reduces the leftover integral to $\int x^{2}\,dx$ .

Does x^2 ln x need integration by parts twice?

No. One pass removes the logarithm completely. The two-pass version of this shape is $\int x^{2}e^{x}\,dx$ , where the polynomial is $u$ .