Question Statement

Evaluate the integral:

$\int x^3\ln xdx$

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Background and Explanation

This integral requires integration by parts, a technique for integrating products of functions. When choosing which part to set as $u$, we follow the LIATE priority rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential), selecting $u=\ln x$ because its derivative simplifies the integral.

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Solution

Let $I$ denote the integral:

$I=\int x^3\ln xdx$

We apply integration by parts using the formula $\int udv=uv-\int vdu$.

Step 1: Choose the parts

• Let $u=\ln x$, which gives $du=\frac{1}{x}dx$
• Let $dv=x^{3}dx$, which gives $v=\int x^{3}dx=\frac{x^4}{4}$

Step 2: Apply the integration by parts formula Substituting into $\int udv=u\cdot v-\int vdu$:

$I=\ln x\cdot \int x^{3}dx-\int \left(\int x^{3}dx\right)\cdot \frac{d}{dx}(\ln x)dx$

$=\ln x\cdot \frac{x^4}{4}-\int \frac{x^4}{4}\cdot \frac{1}{x}dx$

Step 3: Simplify the remaining integral $=\frac{x^4}{4}\ln x-\frac{1}{4}\int x^{3}dx$

Step 4: Evaluate the final integral using the power rule $=\frac{x^4}{4}\ln x-\frac{1}{4}\cdot \frac{x^4}{4}+c$

Step 5: Simplify the expression $=\frac{x^4}{4}\ln x-\frac{x^4}{16}+c$

This can also be factored as: $=\frac{x^4}{16}(4\ln x-1)+c$

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Key Formulas or Methods Used

• Integration by parts: $\int udv=uv-\int vdu$
• Power rule for integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Derivative of natural logarithm: $\frac{d}{dx}(\ln x)=\frac{1}{x}$

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Summary of Steps

1. Identify this as an integration by parts problem with $u=\ln x$ and $dv=x^{3}dx$
2. Compute $du=\frac{1}{x}dx$ and $v=\frac{x^4}{4}$
3. Apply the formula: $\frac{x^4}{4}\ln x-\int \frac{x^4}{4}\cdot \frac{1}{x}dx$
4. Simplify the remaining integral to $\frac{1}{4}\int x^{3}dx$
5. Evaluate to obtain $\frac{x^4}{16}$
6. Combine terms and add the constant of integration: $\frac{x^4}{4}\ln x-\frac{x^4}{16}+C$