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integrate xlnx

Understand the Problem

The question is asking for the integral of the function xln(x) with respect to x. This involves applying integration techniques, possibly using integration by parts.

Answer

The integral is $$ \frac{x^2 \ln(x)}{2} - \frac{x^2}{4} + C $$
Answer for screen readers

The integral of $x \ln(x)$ with respect to $x$ is:
$$ \int x \ln(x) , dx = \frac{x^2 \ln(x)}{2} - \frac{x^2}{4} + C $$

Steps to Solve

  1. Identify the function and the method
    The function we want to integrate is $x \ln(x)$. Integration by parts is appropriate here, where we can let:

    • $u = \ln(x)$ (which will be differentiated)
    • $dv = x , dx$ (which will be integrated)
  2. Differentiate and integrate
    Now, differentiate $u$ and integrate $dv$:

    • $du = \frac{1}{x} , dx$
    • $v = \frac{x^2}{2}$
  3. Apply the integration by parts formula
    Using the integration by parts formula, $\int u , dv = uv - \int v , du$:
    $$ \int x \ln(x) , dx = \ln(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} , dx $$
    This simplifies to:
    $$ = \frac{x^2 \ln(x)}{2} - \int \frac{x}{2} , dx $$

  4. Integrate the remaining integral
    Now we integrate the remaining expression:
    $$ \int \frac{x}{2} , dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} $$

  5. Put it all together
    Now substituting back into our equation gives:
    $$ \int x \ln(x) , dx = \frac{x^2 \ln(x)}{2} - \frac{x^2}{4} + C $$
    where $C$ is the constant of integration.

The integral of $x \ln(x)$ with respect to $x$ is:
$$ \int x \ln(x) , dx = \frac{x^2 \ln(x)}{2} - \frac{x^2}{4} + C $$

More Information

This result uses the technique of integration by parts, which is often employed for products of polynomial and logarithmic functions. It exemplifies how complex integration can be broken down into manageable steps using strategic substitutions.

Tips

  • Forgetting to include the constant of integration $C$ at the end.
  • Misapplying the integration by parts formula. Always ensure you clearly identify $u$ and $dv$.
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