Integration of log(x)
Understand the Problem
The question is asking for the integral of the logarithmic function log(x). To solve this, we will typically use integration by parts, where we can set u = log(x) and dv = dx.
Answer
$$ \int \log(x) \, dx = x \log(x) - x + C $$
Answer for screen readers
$$ \int \log(x) , dx = x \log(x) - x + C $$
Steps to Solve
- Identify the parts for integration by parts
We will use the integration by parts formula:
$$ \int u , dv = u v - \int v , du $$
Set $u = \log(x)$ and $dv = dx$.
- Differentiate u and integrate dv
Now, we need to find $du$ and $v$.
Differentiate $u$:
$$ du = \frac{1}{x} , dx $$
Integrate $dv$:
$$ v = \int dx = x $$
- Apply the integration by parts formula
Substituting $u$, $du$, $v$, and $dv$ into the integration by parts formula gives us:
$$ \int \log(x) , dx = x \log(x) - \int x \left(\frac{1}{x}\right) , dx $$
- Simplify the integral
The term $\int x \left(\frac{1}{x}\right) , dx$ simplifies to:
$$ \int 1 , dx = x $$
Substituting this back gives:
$$ \int \log(x) , dx = x \log(x) - x + C $$
where $C$ is the constant of integration.
- Final Result
The final result after integrating $\log(x)$ is:
$$ \int \log(x) , dx = x \log(x) - x + C $$
$$ \int \log(x) , dx = x \log(x) - x + C $$
More Information
The result of integrating the logarithmic function $\log(x)$ is crucial in various fields such as calculus and analysis. Knowing how to apply integration by parts allows for solving more complex integrals involving logarithmic functions.
Tips
- Forgetting to include the constant of integration $C$ in the final answer. Always remember to add it, as it's important for indefinite integrals.
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