Antiderivative of ln 2x (Calculus Solution)
For real (x>0), the antiderivative of (\ln(2x)) is (x\ln(2x)-x+C). Use integration by parts, taking (u=\ln(2x)) and (dv=dx). The condition (x>0) keeps the logarithm real, and (C) represents any constant. Differentiate the result to check that it returns the original integrand.
If you are learning integration, this problem can look harder than it is. The main steps are to read the expression correctly, choose a useful method, and check your result. Each step matters: a small change in how you read “ln 2x” can lead to a different answer.
In my work explaining calculus, I often see students try substitution first because they have learned that it can simplify expressions. Here, integration by parts is the direct method. Once you see why it fits, the calculation becomes a short sequence rather than a trick to memorize.
Problem Interpretation and Domain
Before integrating, identify the exact function and where it is defined. In this guide, (\ln(2x)) means the natural logarithm of the product (2x). For a real-valued logarithm, its input must be positive, so (2x>0), which gives the domain (x>0).
The notation can be confusing when parentheses are missing. Compare these two expressions:
- (\ln(2x)) means the logarithm of (2x).
- (x\ln 2), sometimes typed as “ln 2 x,” means (x) times the constant (\ln 2).
These are different integrands, so they have different antiderivatives. Unless parentheses or context say otherwise, the problem here is to find an antiderivative of (\ln(2x)).
The domain restriction also applies to the answer. The expression (x\ln(2x)-x+C) is real-valued for (x>0). In calculus, an antiderivative is a function whose derivative gives the integrand on the interval being considered. Here, that interval is ((0,\infty)).
If your worksheet or calculator uses ambiguous notation, rewrite it with parentheses before starting. Writing (\ln(2x)) makes the intended input clear and helps prevent a notation error from becoming an integration error.
Integration-by-Parts Solution
Integration by parts is a method for integrating a product of functions. Its formula is (\int u\,dv=uv-\int v\,du). For (\ln(2x)), view the integrand as the product (\ln(2x)\cdot 1). Choosing (u=\ln(2x)) and (dv=dx) makes both resulting pieces simple.
Start with the integral:
[ \int \ln(2x)\,dx. ]
Set
[ u=\ln(2x), \qquad dv=dx. ]
Now find (du) and (v). To differentiate (u), use the chain rule: the derivative of (\ln(2x)) is (1/(2x)) times the derivative of (2x), which is (2). Thus,
[ du=\frac{1}{x}\,dx. ]
Since (dv=dx), integrating both sides gives
[ v=x. ]
Put these choices into the integration-by-parts formula:
[ \begin{aligned} \int \ln(2x)\,dx &=uv-\int v\,du\ &=x\ln(2x)-\int x\cdot\frac{1}{x}\,dx\ &=x\ln(2x)-\int 1\,dx\ &=x\ln(2x)-x+C. \end{aligned} ]
The factor (x) cancels with (1/x), leaving the simple integral of (1). The final (C) is the constant of integration. Without it, the expression is one antiderivative, but not the complete family.
| Step | Choice or calculation | Result |
|---|---|---|
| Identify the function | Integrand is (\ln(2x)) | (x>0) |
| Choose (u) | (u=\ln(2x)) | (du=\frac1x\,dx) |
| Choose (dv) | (dv=dx) | (v=x) |
| Apply the formula | (uv-\int v\,du) | (x\ln(2x)-\int 1\,dx) |
| Include the constant | Integrate (1) | (x\ln(2x)-x+C) |
A useful way to remember the choice is to separate the expression into a logarithm and a factor of (1). The logarithm is easy to differentiate, while (1) is easy to integrate. That makes the remaining integral simpler than the one you started with.
Verification and Common Errors
A derivative check confirms an antiderivative: differentiate the proposed answer and compare it with the original integrand. For (F(x)=x\ln(2x)-x+C), the product rule and the chain rule give (F'(x)=\ln(2x)+1-1=\ln(2x)). The check works for (x>0).
Differentiate each term with care. For the product (x\ln(2x)), the product rule gives
[ \frac{d}{dx}\left[x\ln(2x)\right] =\ln(2x)+x\left(\frac{1}{x}\right) =\ln(2x)+1. ]
The derivative of (-x) is (-1), and the derivative of a constant is (0). So,
[ \frac{d}{dx}\left[x\ln(2x)-x+C\right] =\ln(2x)+1-1 =\ln(2x). ]
This check catches a common mistake: leaving out the (-x). If you wrote only (x\ln(2x)+C), its derivative would be (\ln(2x)+1), not (\ln(2x)).
Other errors to watch for include:
- Misreading the notation: (\ln(2x)) is not the same as (x\ln 2).
- Missing the chain-rule factor: Differentiating (\ln(2x)) gives (1/x), not (1/(2x)).
- Dropping the integration constant: An indefinite integral needs (+C).
- Ignoring the domain: The real-valued formula applies where (2x>0), so (x>0).
The integration constant does not change the derivative. For example, (x\ln(2x)-x), (x\ln(2x)-x+4), and (x\ln(2x)-x-12) all have the same derivative. They are members of the same family of antiderivatives.
Related Calculus Skills and Applications
This integral connects three core skills: the product rule, the chain rule, and integration by parts. It also shows how an indefinite integral represents a family of functions. The same answer can be checked or used in a definite integral, as long as the input values stay in the domain (x>0).
One helpful comparison is the antiderivative of (\ln x):
[ \int \ln x\,dx=x\ln x-x+C. ]
For (\ln(2x)), the result is (x\ln(2x)-x+C). You can also use the logarithm rule (\ln(2x)=\ln 2+\ln x) for (x>0). Integrating the two terms gives (x\ln 2+x\ln x-x+C), which matches the answer because (\ln 2+\ln x=\ln(2x)).
| Expression to integrate | Interpretation | Antiderivative |
|---|---|---|
| (\ln(2x)) | Logarithm of the product (2x) | (x\ln(2x)-x+C) |
| (x\ln 2) | (x) times the constant (\ln 2) | (\frac{x^2}{2}\ln 2+C) |
| (\ln x) | Logarithm of (x) | (x\ln x-x+C) |
You can also use the result to evaluate a definite integral. For (0<a<b), the fundamental theorem of calculus gives
[ \int_a^b \ln(2x)\,dx =\left[x\ln(2x)-x\right]_a^b. ]
That means you calculate the antiderivative at (b), then subtract its value at (a). The condition (0<a0), where the real logarithm is defined.
For example, from (x=1) to (x=2),
[ \int_1^2 \ln(2x)\,dx =\left[x\ln(2x)-x\right]_1^2 =2\ln 4-2-(\ln 2-1) =3\ln 2-1. ]
The result is a signed area: portions of the graph above the (x)-axis count as positive, and portions below it count as negative. The antiderivative is the tool used to compute that net area.
Conclusion and FAQ
The central result is (\int\ln(2x)\,dx=x\ln(2x)-x+C), for (x>0). The dependable route is to confirm the notation, use integration by parts, include the constant, and differentiate the answer. These checks make the method easier to apply to similar logarithmic integrals.
Frequently Asked Questions
What is the antiderivative of (\ln(2x))?
For (x>0), it is (x\ln(2x)-x+C).
Why use integration by parts?
Write the integrand as (\ln(2x)\cdot1). The logarithm differentiates simply, and (1) integrates to (x).
What are (u) and (dv) in this problem?
Choose (u=\ln(2x)) and (dv=dx). Then (du=(1/x)\,dx) and (v=x).
What is the domain of (\ln(2x))?
For real values, its input must be positive: (2x>0). Therefore, (x>0).
Is (\ln(2x)) the same as (x\ln 2)?
No. The first means the logarithm of (2x). The second means (x) times the constant (\ln 2).
Why does the answer include (-x)?
Integration by parts leaves (-\int 1\,dx), which equals (-x). Without it, differentiating the answer gives an extra (1).
Why must I write (+C)?
An indefinite integral represents all antiderivatives. They differ by a constant, so (+C) records the full family.
How can I check the result?
Differentiate (x\ln(2x)-x+C). The derivative simplifies to (\ln(2x)+1-1=\ln(2x)).
What if the problem means (x\ln 2)?
Then (\ln 2) is a constant, and the antiderivative is (\frac{x^2}{2}\ln 2+C).
Can I use the result for a definite integral?
Yes. Evaluate (x\ln(2x)-x) at the upper endpoint and subtract its value at the lower endpoint, with both endpoints positive.
(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)
