Evaluate the integral: I = integral ln x dx

Evaluate the integral: $$ I = \int \ln x\,dx. $$

Explanation of Method

Treating $\ln x$ as a product $\ln x \cdot 1$ and applying integration by parts with $u=\ln x$, $dv=dx$ turns the integral into a simple polynomial-minus-constant form.

Choose $u=\ln x$, $dv=dx$.
Differentiate $u$ and integrate $dv$.
Apply the integration by parts formula.

Step 1 · Set up integration by parts

Recall $\int u\,dv = uv - \int v\,du$. Let:

$$ u = \ln x \quad \Rightarrow \quad du = \frac{dx}{x}, \qquad dv = dx \quad \Rightarrow \quad v = x. $$

Step 2 · Apply the formula

$$ I = x\ln x - \int x\cdot\frac{dx}{x} = x\ln x - \int dx. $$

Step 3 · Integrate

Since $\int dx = x$:

$$ I = x\ln x - x + C. $$

Summary Table

Quantity	Value
$u$	$\ln x$
$dv$	$dx$
Antiderivative	$x\ln x - x$

Additional Notes

The domain requires $x>0$, where $\ln x$ is defined.
This is the base case for the whole family of $x^n\ln x$-type integrals that follow (Problems 143–145): choosing $u=\ln x$ so that its derivative $1/x$ cancels a power of $x$ from $v$ is the standard integration-by-parts strategy whenever a logarithm is multiplied by a polynomial or constant.
The result can be checked by differentiating $x\ln x-x$ and confirming it returns $\ln x$.

Result

The solution to the integral is:

$$ I = x\ln x - x + C. $$