Integral Of Dx Xlnx: The Step That Unlocks It
Integral of dx xlnx Explained with One Reliable Method
The integral of xlnx is $$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$, and the cleanest way to get it is integration by parts. This result is consistent with standard calculus references, which also show that $$\int \ln x\,dx = x\ln x - x + C$$ and use the same parts identity $$\int u\,dv = uv - \int v\,du$$.
Method
Use integration by parts with $$u=\ln x$$ and $$dv=x\,dx$$. Then $$du=\frac{1}{x}dx$$ and $$v=\frac{x^2}{2}$$, so the formula becomes $$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot \frac{1}{x}\,dx$$.
- Set $$u=\ln x$$, so $$du=\frac{1}{x}dx$$.
- Set $$dv=x\,dx$$, so $$v=\frac{x^2}{2}$$.
- Apply $$\int u\,dv = uv - \int v\,du$$.
- Simplify the remaining integral to $$\frac{1}{2}\int x\,dx$$.
- Finish with $$\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$.
Worked derivation
Starting from $$\int x\ln x\,dx$$, the parts setup gives $$\frac{x^2}{2}\ln x - \frac{1}{2}\int x\,dx$$. Evaluating the last integral yields $$\frac{x^2}{2}\ln x - \frac{1}{2}\cdot \frac{x^2}{2} + C$$, which simplifies to $$\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$.
The key move is to let the logarithm be $$u$$, because its derivative becomes simpler while $$x\,dx$$ is easy to integrate.
Why this works
The logarithm is ideal for $$u$$ because differentiation turns $$\ln x$$ into $$\frac{1}{x}$$, which cancels neatly against the $$x$$ in the product. That cancellation is what makes the remaining integral much easier than the original one.
| Step | Expression | Result |
|---|---|---|
| Choose $$u$$ | $$u=\ln x$$ | $$du=\frac{1}{x}dx$$ |
| Choose $$dv$$ | $$dv=x\,dx$$ | $$v=\frac{x^2}{2}$$ |
| Apply parts | $$\int u\,dv=uv-\int v\,du$$ | $$\frac{x^2}{2}\ln x-\int \frac{x}{2}\,dx$$ |
| Integrate remainder | $$\frac{1}{2}\int x\,dx$$ | $$\frac{x^2}{4}$$ |
| Final answer | $$\int x\ln x\,dx$$ | $$\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$$ |
Common errors
A frequent mistake is choosing $$u=x$$ and $$dv=\ln x\,dx$$, which usually makes the algebra less efficient. Another error is forgetting the constant of integration $$C$$, or dropping the factor of $$\frac{1}{2}$$ after integrating $$x$$.
- Do not forget $$C$$.
- Do not integrate $$\ln x$$ directly without parts.
- Do not lose the factor from $$\int x\,dx = \frac{x^2}{2}$$.
Check by differentiation
Differentiate $$\frac{x^2}{2}\ln x - \frac{x^2}{4}$$ to verify the answer. Using the product rule on $$\frac{x^2}{2}\ln x$$ gives $$x\ln x + \frac{x}{2}$$, and differentiating $$-\frac{x^2}{4}$$ gives $$-\frac{x}{2}$$, so the middle terms cancel and leave $$x\ln x$$.
Practical takeaway
The integral of xlnx has one straightforward solution: use parts, simplify the cancellation, and finish with $$\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$. For students, this is a strong example of how a smart choice of $$u$$ can turn a difficult product into a routine calculation.
Helpful tips and tricks for Integral Of Dx Xlnx The Step That Unlocks It
What is the integral of x ln x?
$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$. This follows directly from integration by parts with $$u=\ln x$$ and $$dv=x\,dx$$.
Why choose ln x as u?
Because differentiating $$\ln x$$ produces $$\frac{1}{x}$$, which simplifies the product with $$x$$. That choice is the standard reliable method used in calculus explanations of this integral.
Can this be used for definite integrals?
Yes, the same integration-by-parts setup works for definite integrals by evaluating the boundary term $$uv$$ at the endpoints and then integrating the remaining term. The definite version uses the same formula with limits.
