Integral of x ln x: Answer, Parts Setup, Mistakes

How to integrate x ln x by parts

The integrand is a product of an algebraic factor and a logarithm, and neither one is the derivative of the other, so substitution has nothing to grip. That combination is the signature of integration by parts.

Choose $u = \ln x$. Differentiating it gives $\frac{1}{x}$, which is algebraic, so the logarithm disappears from the problem in one step. That leaves $dv = x\,dx$ and $v = \frac{x^{2}}{2}$.

Substituting into the formula, the new integral simplifies before you integrate it, because $\frac{x^{2}}{2}\cdot\frac{1}{x}$ collapses to $\frac{x}{2}$.

That remaining integral is the power rule, and it contributes $\frac{x^{2}}{4}$.

Why u = ln x and not u = x

Parts only helps when the new integral $\int v\,du$ is easier than the original. Taking $u = x$ and $dv = \ln x\,dx$ is a dead end for a reason worth naming: finding $v$ means integrating $\ln x$, which is itself a parts problem, so you have made the question harder before you have started.

The LIATE ordering encodes this. Logarithmic sits first precisely because logarithms are awkward to antidifferentiate and pleasant to differentiate.

1. Logarithmic, such as $\ln x$
2. Inverse trigonometric, such as $\arctan x$
3. Algebraic, such as $x$ or $x^{2}$
4. Trigonometric, such as $\sin x$
5. Exponential, such as $e^{x}$

The general power case follows the same working, and it is worth knowing because it covers $\int \sqrt{x}\ln x\,dx$ and similar problems without redoing parts.

Setting $n = 1$ reproduces the result on this page. The excluded case $n = -1$ is genuinely different: $\int \frac{\ln x}{x}\,dx$ is a substitution with $u = \ln x$, giving $\frac{\left(\ln x\right)^{2}}{2} + C$, and no parts is needed.

Where the integral of x ln x shows up on the AP exam

Integration by parts is Topic 6.11 in Unit 6, marked BC only in the CED, so $\int x\ln x\,dx$ can appear on BC but not on AB. Unit 6 carries a weighting of 15 to 20 percent on both exams, and Topic 6.14 is where the technique choice itself is assessed.

Definite versions from $1$ to some upper limit are the common form, because $\ln 1 = 0$ kills the logarithmic term at the lower endpoint and leaves clean arithmetic.

That evaluates to about $2.097$. Note the lower endpoint contributes $+\frac{1}{4}$ because you are subtracting $-\frac{1}{4}$, which is the sign slip most often made here.

The integrand is also a standard improper integral example on BC, since $x\ln x \to 0$ as $x \to 0^{+}$ even though $\ln x$ diverges there. The $x$ factor wins, which is a useful instance of the growth comparison BC expects you to know.

Common mistakes with the integral of x ln x

• Multiplying antiderivatives to get $\frac{x^{2}}{2}\left(x\ln x - x\right)$. There is no product rule for integrals, and differentiating that expression does not return $x\ln x$.
• Choosing $u = x$. It forces you to antidifferentiate $\ln x$ to find $v$, which is a harder problem than the one you were given.
• Dropping the $\frac{1}{4}$ and writing $\frac{x^{2}}{2}\ln x - \frac{x^{2}}{2}$. The leftover integral is $\int \frac{x}{2}\,dx = \frac{x^{2}}{4}$, and halving happens once in $v$ and again in the power rule.
• Confusing this with $\int \frac{\ln x}{x}\,dx$. That one is a substitution, not parts, and its answer is $\frac{\left(\ln x\right)^{2}}{2} + C$.
• Ignoring the domain. $\ln x$ requires $x > 0$, so a definite integral with a negative endpoint does not exist.