Question Statement

Evaluate the integral:

$\int \ln xdx$

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Background and Explanation

This problem requires integration by parts, a technique used to integrate products of functions. This method is particularly effective when one function becomes simpler upon differentiation (like logarithmic functions) while the other is easily integrated (like constants).

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Solution

To evaluate this integral, we recognize that $\ln x$ is not easily integrated directly, but simplifies when differentiated. We rewrite the integrand as a product with the constant function $1$:

$I=\int \ln x\cdot 1dx$

We apply integration by parts using the formula:

$\int uvdx=u\int vdx-\int \left(\frac{d}{dx}(u)\int vdx\right)dx$

Step 1: Choose the functions for $u$ and $v$:

• Let $u=\ln x$ (this simplifies when differentiated)
• Let $v=1$ (this is easily integrated)

Step 2: Compute the required components:

• $\int vdx=\int 1dx=x$
• $\frac{d}{dx}(u)=\frac{d}{dx}(\ln x)=\frac{1}{x}$

Step 3: Substitute into the integration by parts formula:

$\begin{array}{rl}I& =\ln x\cdot \int 1dx-\int \left(\int 1dx\right)\cdot \frac{d}{dx}(\ln x)dx\\ & =\ln x\cdot x-\int x\cdot \frac{1}{x}dx\\ & =x\ln x-\int 1dx\\ & =x\ln x-x+c\end{array}$

Therefore, the solution is:

$\int \ln xdx=x\ln x-x+c$

where $c$ is the constant of integration.

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Key Formulas or Methods Used

• Integration by parts: $\int uvdx=u\int vdx-\int \left(\frac{du}{dx}\int vdx\right)dx$ (or equivalently $\int udv=uv-\int vdu$)
• Derivative of natural logarithm: $\frac{d}{dx}(\ln x)=\frac{1}{x}$
• Basic integral: $\int 1dx=x$

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Summary of Steps

1. Rewrite the integral as $\int \ln x\cdot 1dx$ to prepare for integration by parts
2. Select $u=\ln x$ and $v=1$ (following the LIATE priority rule: Logarithmic before Algebraic)
3. Calculate $\int vdx=x$ and $\frac{du}{dx}=\frac{1}{x}$
4. Apply the integration by parts formula: $u\int vdx-\int \left(\frac{du}{dx}\int vdx\right)dx$
5. Simplify: $x\ln x-\int x\cdot \frac{1}{x}dx=x\ln x-\int 1dx$
6. Complete the integration: $x\ln x-x+c$