1 Lnx Integration: A Subtle Challenge Worth Unpacking
The integral of $$\ln(x)$$ is $$x\ln(x)-x+C$$, found by integration by parts, where $$u=\ln(x)$$ and $$dv=dx$$.
Method
The key step is to treat $$\ln(x)$$ as $$u$$ and $$1$$ as $$dv$$, because differentiating $$\ln(x)$$ simplifies the expression while integrating $$dx$$ is immediate.
Workthrough
- Set $$u=\ln(x)$$, so $$du=\frac{1}{x}dx$$.
- Set $$dv=dx$$, so $$v=x$$.
- Apply integration by parts: $$\int u\,dv=uv-\int v\,du$$.
- Substitute: $$\int \ln(x)\,dx=x\ln(x)-\int x\cdot \frac{1}{x}dx$$.
- Simplify to $$x\ln(x)-\int 1\,dx=x\ln(x)-x+C$$.
Why It Works
This is a classic example of using integration by parts to convert a function that is awkward to integrate directly into one that becomes elementary after differentiation and simplification.
Formula Table
| Expression | Result |
|---|---|
| $$\int \ln(x)\,dx$$ | $$x\ln(x)-x+C$$ |
| $$\frac{d}{dx}[x\ln(x)-x]$$ | $$\ln(x)$$ |
Common Mistake
A frequent error is to try to integrate $$\ln(x)$$ directly without parts; the simplest route is to rewrite it as $$\ln(x)\cdot 1$$ and use the integration-by-parts identity.
What are the most common questions about 1 Lnx Integration A Subtle Challenge Worth Unpacking?
What is the integral of $$\ln(x)$$?
The antiderivative is $$x\ln(x)-x+C$$.
Which method should be used?
Use integration by parts, with $$u=\ln(x)$$ and $$dv=dx$$.
Does the result need an absolute value?
For real-valued calculus, $$\ln(x)$$ is defined for $$x>0$$, so the standard antiderivative is written as $$x\ln(x)-x+C$$ on that domain.
