Introduction
The integral (\displaystyle \int 2x^{2}\ln x ,dx) is a classic example that combines a polynomial factor with a logarithmic function. Plus, mastering this type of integral not only strengthens your calculus toolkit but also deepens your understanding of integration by parts, a method that appears repeatedly in physics, engineering, and economics. In this article we will explore the step‑by‑step solution, discuss the underlying theory, present alternative approaches, and answer common questions that students often encounter when tackling (\int 2x^{2}\ln x ,dx). By the end of the reading, you will be able to solve this integral confidently and apply the same techniques to a wide range of similar problems Practical, not theoretical..
Why Integration by Parts?
When an integrand is the product of two functions—one that becomes simpler when differentiated and another that remains manageable when integrated—integration by parts is the tool of choice. The formula
[ \int u,dv = uv - \int v,du ]
mirrors the product rule for differentiation and turns a difficult integral into a combination of easier ones. In (\displaystyle 2x^{2}\ln x) we have:
- ( \ln x ) – a logarithmic function that simplifies dramatically when differentiated ((\frac{d}{dx}\ln x = \frac{1}{x})).
- ( 2x^{2} ) – a polynomial that is trivial to integrate ((\int 2x^{2}dx = \frac{2}{3}x^{3})).
Choosing (u = \ln x) and (dv = 2x^{2},dx) therefore satisfies the “differentiate‑the‑log, integrate‑the‑polynomial” strategy That alone is useful..
Step‑by‑Step Solution
1. Identify (u) and (dv)
[ u = \ln x \qquad\Longrightarrow\qquad du = \frac{1}{x},dx ]
[ dv = 2x^{2},dx \qquad\Longrightarrow\qquad v = \int 2x^{2},dx = \frac{2}{3}x^{3} ]
2. Apply the integration‑by‑parts formula
[ \int 2x^{2}\ln x ,dx = uv - \int v,du ]
Substituting the expressions for (u, v, du):
[ \int 2x^{2}\ln x ,dx = \left(\ln x\right)!\left(\frac{2}{3}x^{3}\right) - \int \frac{2}{3}x^{3}\cdot\frac{1}{x},dx ]
Simplify the remaining integral:
[ \int \frac{2}{3}x^{3}\cdot\frac{1}{x},dx = \int \frac{2}{3}x^{2},dx = \frac{2}{3}\int x^{2},dx = \frac{2}{3}\cdot\frac{x^{3}}{3}= \frac{2}{9}x^{3} ]
3. Assemble the result
[ \int 2x^{2}\ln x ,dx = \frac{2}{3}x^{3}\ln x - \frac{2}{9}x^{3} + C ]
where (C) denotes the constant of integration.
4. Factor for a cleaner expression
[ \boxed{\displaystyle \int 2x^{2}\ln x ,dx = \frac{2x^{3}}{9}\bigl(3\ln x - 1\bigr) + C} ]
Both forms are mathematically equivalent; the factored version often looks tidier in textbooks and exam solutions Worth keeping that in mind..
Verifying the Result
A quick differentiation confirms the answer:
[ \frac{d}{dx}\Bigl[\frac{2x^{3}}{9}\bigl(3\ln x - 1\bigr)\Bigr] = \frac{2}{9}\Bigl[3x^{2}(3\ln x - 1) + x^{3}\cdot\frac{3}{x}\Bigr] = \frac{2}{9}\Bigl[9x^{2}\ln x - 3x^{2} + 3x^{2}\Bigr] = 2x^{2}\ln x. ]
The derivative matches the original integrand, confirming the correctness of the antiderivative.
Alternative Approaches
1. Substitution Before Integration by Parts
Sometimes students wonder whether a substitution such as (x = e^{t}) could simplify the integral. Setting (x = e^{t}) gives (dx = e^{t}dt) and (\ln x = t). The integral becomes:
[ \int 2(e^{t})^{2},t,e^{t},dt = \int 2e^{3t}t,dt. ]
Now the problem reduces to (\int t e^{3t},dt), which still requires integration by parts (but with a simpler exponential factor). The final answer, after back‑substituting (t = \ln x), coincides with the result obtained earlier. This route illustrates that integration by parts is inevitable; the substitution only reshapes the algebraic landscape It's one of those things that adds up..
2. Using a Tabular Method
For products of a polynomial and a logarithmic (or trigonometric) function, the tabular integration technique streamlines the process:
| Derivative of (u=\ln x) | Integral of (dv=2x^{2}dx) |
|---|---|
| (\displaystyle \frac{1}{x}) | (\displaystyle \frac{2}{3}x^{3}) |
| (\displaystyle -\frac{1}{x^{2}}) | (\displaystyle \frac{2}{12}x^{4}) (not needed) |
We stop after the first derivative because the next derivative would introduce a negative power that complicates the integral. Multiplying diagonally and alternating signs reproduces the same expression (\frac{2}{3}x^{3}\ln x - \frac{2}{9}x^{3}). The tabular method is especially handy when the polynomial degree is higher (e.g., (x^{5}\ln x)), saving repetitive integration‑by‑parts steps.
Scientific Explanation: Why Does Integration by Parts Work?
Integration by parts originates from the product rule for differentiation:
[ \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}. ]
Rearranging terms and integrating both sides yields
[ \int u,dv = uv - \int v,du. ]
In essence, the technique transfers differentiation from one factor to the other. Because of that, in (\int 2x^{2}\ln x,dx), moving the derivative from the logarithm (which becomes (1/x)) to the polynomial (which becomes a higher‑degree term) reduces the overall algebraic complexity. The method is a direct consequence of the linearity of the integral and the product rule, guaranteeing its validity for any pair of sufficiently smooth functions.
Frequently Asked Questions
Q1: Can I choose (u = 2x^{2}) and (dv = \ln x,dx) instead?
Yes, you can, but the resulting integral becomes more cumbersome. Differentiating (2x^{2}) yields (4x), while integrating (\ln x) requires its own integration by parts, leading to a nested process. The standard heuristic—differentiate the logarithm, integrate the polynomial—produces the shortest path Took long enough..
Q2: What if the limits of integration are given, e.g., (\int_{1}^{e} 2x^{2}\ln x,dx)?
Apply the antiderivative we derived and evaluate at the bounds:
[ \left[\frac{2x^{3}}{9}\bigl(3\ln x - 1\bigr)\right]_{1}^{e} = \frac{2e^{3}}{9}\bigl(3\cdot1 - 1\bigr) - \frac{2}{9}\bigl(3\cdot0 - 1\bigr) = \frac{2e^{3}}{9}\cdot2 + \frac{2}{9} = \frac{4e^{3}+2}{9}. ]
The definite integral evaluates cleanly because the antiderivative is explicit.
Q3: Is there a way to avoid the constant of integration (C) in indefinite integrals?
No. The constant (C) reflects the family of all antiderivatives differing by a constant. Whenever you solve an indefinite integral, always append (+C) unless a specific initial condition is provided to determine its value It's one of those things that adds up..
Q4: How does this integral relate to real‑world applications?
Expressions of the form (x^{n}\ln x) appear in entropy calculations in thermodynamics, information theory, and growth models where a logarithmic correction modifies a power‑law behavior. Computing the area under such curves often requires integrals like (\int 2x^{2}\ln x,dx) Easy to understand, harder to ignore..
Q5: What if the integrand were (2x^{2}(\ln x)^{2})?
You would still use integration by parts, but you would need to apply the method twice because differentiating ((\ln x)^{2}) once yields (2\ln x / x), which still contains a logarithm. Repeating the process removes the logarithmic factor entirely, leaving a polynomial integral Easy to understand, harder to ignore..
Common Mistakes to Avoid
| Mistake | Why It Happens | How to Fix It |
|---|---|---|
| Forgetting the factor (2) when integrating (2x^{2}) | Over‑reliance on memory of (\int x^{n}) | Write (\int 2x^{2}dx = 2\cdot\frac{x^{3}}{3}) explicitly |
| Mixing up (u) and (dv) | Confusion between “differentiate” and “integrate” roles | Remember the rule: differentiate the log, integrate the polynomial |
| Dropping the constant (C) in an indefinite integral | Belief that the constant is “optional” | Always add (+C) unless a boundary condition is supplied |
| Sign errors when applying (uv - \int v,du) | Misreading the minus sign | Write the formula on a scrap paper before substituting values |
Conclusion
The integral (\displaystyle \int 2x^{2}\ln x ,dx) serves as a perfect illustration of how integration by parts transforms a seemingly complex product into a manageable expression. By selecting (u = \ln x) and (dv = 2x^{2}dx), the problem reduces to a simple polynomial integral, leading to the compact antiderivative
[ \frac{2x^{3}}{9}\bigl(3\ln x - 1\bigr) + C. ]
Understanding the rationale behind the choice of (u) and (dv), verifying the result through differentiation, and recognizing alternative strategies such as substitution or the tabular method enriches your problem‑solving repertoire. Beyond that, the techniques discussed here extend far beyond this single example—they are foundational tools for tackling a broad class of integrals that blend logarithmic, exponential, and polynomial components Not complicated — just consistent..
Keep practicing with variations (different powers of (x), higher powers of (\ln x), or added constants) to cement the method. But as you encounter more complex integrals in physics, engineering, or data science, the confidence gained from mastering (\int 2x^{2}\ln x ,dx) will make the journey smoother and more rewarding. Happy integrating!
Extending the Idea: Powers and Products
When the polynomial part has a higher degree, the same pattern holds: each time you integrate the polynomial you raise its power by one and divide by the new exponent, while the logarithmic factor remains untouched until the final subtraction step. Here's one way to look at it:
[ \int 4x^{5}\ln x,dx ]
requires the same two‑step integration‑by‑parts routine:
| Step | Choice | Result |
|---|---|---|
| 1 | (u=\ln x,; dv=4x^{5}dx) | (du=\frac{1}{x}dx,; v=\frac{4x^{6}}{6}=\frac{2x^{6}}{3}) |
| 2 | Apply (uv-\int v,du) | (\displaystyle \frac{2x^{6}}{3}\ln x-\int\frac{2x^{6}}{3}\cdot\frac{1}{x}dx = \frac{2x^{6}}{3}\ln x-\frac{2}{3}\int x^{5}dx) |
| 3 | Integrate the remaining monomial | (\frac{2x^{6}}{3}\ln x-\frac{2}{3}\cdot\frac{x^{6}}{6}+C = \frac{2x^{6}}{3}\Bigl(\ln x-\frac{1}{6}\Bigr)+C). |
The final answer mirrors the structure we obtained for the original problem: a polynomial factor multiplied by a linear combination of (\ln x) and a constant. In general,
[ \int a,x^{n}\ln x,dx = a\frac{x^{n+1}}{n+1}\Bigl(\ln x-\frac{1}{n+1}\Bigr)+C, ]
where (a) is any constant and (n\neq -1). This compact formula can be derived directly from the integration‑by‑parts steps shown above and provides a quick shortcut for many textbook exercises Turns out it matters..
When Multiple Logarithms Appear
If the integrand contains a higher power of the logarithm, such as ((\ln x)^{k}) with (k\ge 2), each differentiation of ((\ln x)^{k}) reduces the exponent by one while introducing a factor of (k/x). Because of this, you will need to perform integration by parts k times. The process generates a finite series:
[ \int x^{n}(\ln x)^{k}dx = \frac{x^{n+1}}{n+1}\sum_{j=0}^{k}(-1)^{j}\frac{k!}{(k-j)!}\frac{(\ln x)^{k-j}}{(n+1)^{j}} + C Most people skip this — try not to..
For the specific case (k=2) and (n=2),
[ \int 2x^{2}(\ln x)^{2}dx = \frac{2x^{3}}{3}\Bigl[(\ln x)^{2}-\frac{2\ln x}{3}+\frac{2}{9}\Bigr]+C, ]
which you can verify by differentiating the right‑hand side. Which means this formula underscores how the repeated application of the product rule in reverse (i. e., integration by parts) systematically peels away the logarithmic layers.
Tabular Integration: A Faster Route
When you anticipate many repetitions of the same pattern—polynomial times logarithm or polynomial times exponential—the tabular method streamlines the bookkeeping. The table for (\int 2x^{2}\ln x,dx) looks like this:
| (u) (differentiate) | (dv) (integrate) |
|---|---|
| (\ln x) | (2x^{2}) → (\frac{2x^{3}}{3}) |
| (\frac{1}{x}) | (\frac{2x^{3}}{3}) → (\frac{2x^{2}}{3}) |
| (-\frac{1}{x^{2}}) | (\frac{2x^{2}}{3}) → (\frac{4x}{9}) |
| … | … |
You then multiply diagonally, alternating signs, and stop when the derivative column becomes zero. For our simple integral only the first two rows survive, reproducing the result instantly:
[ \frac{2x^{3}}{3}\ln x - \frac{2x^{3}}{9} + C. ]
The tabular technique is especially handy in exams where time is limited; once you internalize the pattern, you can write down the answer without a single integral sign remaining.
Numerical Checks and Software Verification
Even after a clean symbolic derivation, it is good practice to verify the antiderivative numerically, especially when the expression will be used in a larger model. A quick sanity check in Python (or any CAS) looks like this:
import sympy as sp
x = sp.symbols('x', positive=True)
F = (2*x**3/9)*(3*sp.log(x) - 1) # our antiderivative
sp.diff(F, x).simplify()
The output should be 2*x**2*log(x), confirming that no algebraic slip occurred. Similar checks can be performed in Mathematica, Maple, or even a graphing calculator Small thing, real impact..
Real‑World Contexts Where This Integral Appears
| Field | Typical Appearance |
|---|---|
| Thermodynamics | Entropy of an ideal gas: (S \propto \int \ln(V) dV) after a change of variables leads to terms like (x^{2}\ln x). Plus, |
| Information Theory | Expected code length for a distribution with a polynomial tail introduces (\int p(x)\ln p(x)dx) where (p(x)) may be proportional to (x^{2}). On the flip side, |
| Econometrics | Production functions of the form (Y = A K^{\alpha}L^{\beta}(\ln K)^{\gamma}) generate integrals of (K^{\alpha}\ln K) when aggregating over capital. |
| Computer Graphics | When computing moments of a radial density that includes a logarithmic term, the integral of (r^{2}\ln r) over a disc surfaces. |
In each of these scenarios, the analytic antiderivative simplifies the analysis, allowing closed‑form expressions for quantities like total entropy, average code length, or cumulative production.
Final Thoughts
The integral (\displaystyle \int 2x^{2}\ln x,dx) may look intimidating at first glance, but it is a textbook example of how a strategic choice of (u) and (dv) reduces a mixed algebraic‑logarithmic expression to elementary pieces. By:
- Choosing (u=\ln x) (the part that simplifies under differentiation) and (dv=2x^{2}dx) (the part that integrates cleanly);
- Applying the product‑rule reversal (uv-\int v,du);
- Carrying out the remaining polynomial integral; and
- Verifying the result through differentiation or a CAS,
you obtain the compact antiderivative
[ \boxed{\displaystyle \frac{2x^{3}}{9}\bigl(3\ln x-1\bigr)+C }. ]
Beyond this single case, the same pattern extends to any power of (x) multiplied by a logarithm, to higher powers of (\ln x) (with repeated parts), and even to mixed exponential‑logarithmic products. Mastering the underlying logic—differentiate the “hard” part, integrate the “easy” part—equips you with a versatile tool for a wide swath of problems across mathematics, physics, engineering, and data science.
So the next time you encounter a product of a polynomial and a logarithm, remember: a few well‑placed parts, a dash of the integration‑by‑parts formula, and a quick check are all you need to turn a daunting integral into a neat, closed‑form answer. Happy integrating!
A convenient alternative to parts is the substitution (t=\ln x), which turns the integral into a pure polynomial. With (x=e^{t}) and (dx=e^{t}dt),
[ \int 2x^{2}\ln x,dx =\int 2e^{2t},t,e^{t}dt =\int 2t,e^{3t}dt. ]
Now integrate by parts in the variable (t) (or use the standard formula (\int t e^{at}dt = \frac{e^{at}}{a^{2}}(at-1)+C)). The result simplifies to
[ \frac{2e^{3t}}{9},(3t-1)+C. ]
Re‑expressing (t) as (\ln x) and (e^{3t}=x^{3}) yields exactly the same antiderivative obtained earlier:
[ \boxed{\frac{2x^{3}}{9}\bigl(3\ln x-1\bigr)+C}. ]
A quick differentiation confirms the correctness:
[ \frac{d}{dx}!\left[\frac{2x^{3}}{9}\bigl(3\ln x-1\bigr)\right] =\frac{2}{9}\bigl(9x^{2}\ln x+3x^{2}-x^{2}\bigr) =2x^{2}\ln x. ]
In symbolic environments this check is often written as 2*x**2*log(x), a compact way to verify that no algebraic slip occurred. The same pattern works for any power (x^{n}) multiplied by a logarithm; the substitution (t=\ln x) reduces the problem to (\int t,e^{(n+1)t}dt), which is elementary.
Beyond the basic case, repeated integration by parts handles higher‑order logarithmic factors. To give you an idea,
[ \int x^{2}\bigl(\ln x\bigr)^{2},dx ]
requires two successive applications of the parts formula, each time lowering the power of the logarithm until a polynomial remains. The general formula after (k) repetitions is
[ \int x^{n}(\ln x)^{k},dx =\frac{x^{n+1}}{n+1}\sum_{j=0}^{k}(-1)^{j}\frac{k!}{(k-j)!}\frac{(\ln x)^{,k-j}}{(n+1)^{j}}+C. ]
Such expressions appear in the computation of moments for distributions whose density contains a logarithmic modifier, in the analysis of algorithms where a cost function grows like (n^{2}\log n), and in the derivation of Fisher information for parametric models with polynomial‑logarithmic likelihoods.
The versatility of integration by parts, together with the simplicity of the substitution route, makes the integral (\int 2x^{2}\ln x,dx) a prototypical example of how a strategic choice of “u” and “dv” (or an appropriate change of variables) can transform a seemingly layered expression into a clean, closed‑form result. Mastery of these techniques equips the reader to tackle a broad spectrum of problems that arise in physics, statistics, engineering, and computer science, where polynomial‑logarithmic products are a common occurrence.
Conclusion
By selecting the logarithmic factor for differentiation and the polynomial for integration, or by converting to a logarithmic variable, the integral resolves neatly to (\frac{2x^{3}}{9}(3\ln x-1)+C), and a straightforward derivative check confirms the absence of any algebraic error. This exemplifies the power of systematic methods in calculus and underscores their relevance across diverse scientific and engineering domains.
