MathSolver.site

Integral of x ln x: Integration by Parts

The integral ∫x ln x dx is the classic first meeting with integration by parts for most students. It looks intimidating because no basic rule (power rule, substitution) fits: the integrand is a product of an algebraic piece (x) and a logarithm (ln x), and neither piece differentiates into the other in a helpful way.

The way out is integration by parts, which is just the product rule for derivatives run in reverse. The hard part is not the formula; it is choosing which part of the integrand becomes u and which becomes dv. Choose badly and the new integral is worse than the old one. Choose well and the logarithm vanishes, leaving a trivial power integral. This worked example shows the good choice, why it is the good choice, and how to check the answer.

We will show that ∫x ln x dx = (x²/2)·ln x − x²/4 + C.

Worked solution

Step 1: Recall the integration-by-parts formula.

∫u dv = u·v − ∫v du

You split the integrand into two factors: one you will differentiate (u) and one you will integrate (dv). Then you trade the original integral for a new one, ∫v du, which had better be easier.

Step 2: Choose u and dv with ILATE.

ILATE is a priority list for picking u: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. Whichever type appears first on the list becomes u.

Our integrand has two factors: ln x (Logarithmic) and x (Algebraic). Logarithmic beats Algebraic on the ILATE list, so:

  • u = ln x (differentiate this)
  • dv = x dx (integrate this)

Why is this the right call? Differentiating ln x gives 1/x, which is simpler than ln x. Integrating x gives x²/2, which is no worse than x. The logarithm, the awkward piece, is the one we differentiate away. If we had chosen u = x instead, we would end up integrating ln x, which is the harder direction.

Step 3: Compute du and v.

  • u = ln x → du = (1/x) dx
  • dv = x dx → v = ∫x dx = x²/2

Step 4: Plug into the formula.

∫x ln x dx = u·v − ∫v du
= (ln x)(x²/2) − ∫(x²/2)(1/x) dx
= (x²/2)·ln x − ∫(x²/2)·(1/x) dx

Step 5: Simplify the new integral.

(x²/2)·(1/x) = x/2, because one power of x cancels:

∫(x²/2)·(1/x) dx = ∫(x/2) dx = (1/2)·(x²/2) = x²/4

Step 6: Write the final answer.

∫x ln x dx = (x²/2)·ln x − x²/4 + C ✓

Step 7: Check (never skip): differentiate the answer.

If our antiderivative is right, its derivative must be the original integrand x ln x. Differentiate term by term. For (x²/2)·ln x, use the product rule:

  • d/dx[(x²/2)·ln x] = x·ln x + (x²/2)·(1/x) = x·ln x + x/2
  • d/dx[−x²/4] = −x/2
  • d/dx[C] = 0

Add them: (x·ln x + x/2) + (−x/2) = x·ln x ✓

The derivative is exactly the integrand we started with. The answer is confirmed.

Why this method works

Integration by parts comes from the product rule: d(uv)/dx = u·dv/dx + v·du/dx. Integrate both sides and rearrange, and you get ∫u dv = uv − ∫v du. It is not a trick; it is a rearranged product rule. It works on ∫x ln x dx because differentiating the logarithm simplifies it (ln x → 1/x) while integrating the algebraic factor keeps it tame (x → x²/2). The new integral, ∫x/2 dx, is one you can do in your sleep. That trade, a hard integral for an easy one, is the whole point of the method.

Common mistakes

  • Picking u = x instead of u = ln x. Then dv = ln x dx, and you must integrate ln x, which itself needs integration by parts. You go in circles. ILATE exists precisely to prevent this: Logarithmic outranks Algebraic, so u = ln x.
  • Forgetting to multiply v and du correctly. The new integral is ∫(x²/2)(1/x) dx, not ∫(x²/2) dx. Students often drop the 1/x from du. Write du = (1/x)dx explicitly before substituting.
  • Dropping the constant C. An indefinite integral always needs + C. Also, in the check step, remember that differentiating kills C, which is why the check still works.

FAQ

Q: What does ILATE stand for, and do I have to memorize it? A: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. It is a priority order for choosing u, and it is worth memorizing because it turns a judgment call into a routine. In ∫x ln x dx, Logarithmic (ln x) beats Algebraic (x), so u = ln x.

Q: Can I check an integration-by-parts answer another way? A: Differentiating the answer, as we did in Step 7, is the gold standard: if d/dx of your result equals the integrand, the antiderivative is correct. It catches sign errors and dropped terms immediately.

Q: Does this method work for ∫x² ln x dx too? A: Yes, with the same choice: u = ln x, dv = x² dx. Then v = x³/3 and the new integral is ∫(x³/3)(1/x) dx = ∫x²/3 dx = x³/9. The answer is (x³/3)ln x − x³/9 + C. The pattern is the same every time the logarithm meets a power of x.

Evaluate integrals with steps on our integral calculator.