Question Statement

Evaluate the integral:

$\int (\ln x{)}^{2}dx$

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Background and Explanation

This problem requires integration by parts, a technique based on the product rule for differentiation. Since $(\ln x{)}^2$ is a logarithmic function that doesn't have an immediate antiderivative, we apply integration by parts twice—first to reduce the power from 2 to 1, and again to eliminate the logarithm entirely.

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Solution

We begin by denoting the integral as $I$ and preparing for integration by parts by introducing a factor of $1$:

$I=\int (\ln x{)}^2\cdot 1dx$

First Application of Integration by Parts

Using the formula $\int udv=uv-\int vdu$, we choose:

• $u=(\ln x{)}^2$ (so that $du$ reduces the power)
• $dv=1dx$ (so $v=x$)

Differentiating $u$: $\frac{d}{dx}(\ln x{)}^2=2(\ln x{)}^{2-1}\cdot \frac{1}{x}=\frac{2\ln x}{x}$

Applying integration by parts:

$\begin{array}{rl}I& =(\ln x{)}^2\cdot x-\int x\cdot \frac{2\ln x}{x}dx\\ & =x(\ln x{)}^2-2\int \ln xdx\end{array}$

Second Application of Integration by Parts

Now we evaluate $\int \ln xdx$ by applying integration by parts again. Rewrite it as $\int \ln x\cdot 1dx$ and choose:

• $u=\ln x$
• $dv=1dx$ (so $v=x$)

Differentiating $u$: $\frac{d}{dx}(\ln x)=\frac{1}{x}$

Applying the formula:

$\begin{array}{rl}\int \ln xdx& =\ln x\cdot x-\int x\cdot \frac{1}{x}dx\\ & =x\ln x-\int 1dx\\ & =x\ln x-x\end{array}$

Substituting Back

Replace $\int \ln xdx$ in our expression for $I$:

$\begin{array}{rl}I& =x(\ln x{)}^2-2\left[x\ln x-x\right]+c\\ & =x(\ln x{)}^2-2x\ln x+2x+c\end{array}$

This can also be written in factored form:

$I=x(\ln x{)}^2-2x(\ln x-1)+c$

Or equivalently:

$I=x\left[(\ln x{)}^2-2\ln x+2\right]+c$

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Key Formulas or Methods Used

• Integration by Parts: $\int udv=uv-\int vdu$ (or equivalently $\int u\cdot vdx=u\int vdx-\int \left(\frac{du}{dx}\int vdx\right)dx$)
• Chain Rule for Differentiation: $\frac{d}{dx}(\ln x{)}^n=n(\ln x{)}^{n-1}\cdot \frac{1}{x}$
• Derivative of Natural Log: $\frac{d}{dx}(\ln x)=\frac{1}{x}$

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Summary of Steps

1. Set up: Write the integral as $I=\int (\ln x{)}^2\cdot 1dx$ to prepare for integration by parts
2. First integration by parts: Choose $u=(\ln x{)}^2$ and $dv=dx$ to obtain $x(\ln x{)}^2-2\int \ln xdx$
3. Second integration by parts: Apply the same method to $\int \ln xdx$ with $u=\ln x$ and $dv=dx$ to get $x\ln x-x$
4. Substitute: Replace the remaining integral to obtain $x(\ln x{)}^2-2(x\ln x-x)+c$
5. Simplify: Expand and factor if desired to reach the final form $x(\ln x{)}^2-2x\ln x+2x+c$