Primitive of Logarithm of x/Proof 1
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Theorem
- $\ds \int \ln x \rd x = x \ln x - x + C$
Proof
With a view to expressing the primitive in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds \ln x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac 1 x\) | Derivative of $\ln x$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds x\) | Primitive of Constant |
Then:
\(\ds \int \ln x \rd x\) | \(=\) | \(\ds x \ln x - \int x \paren {\frac 1 x} \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds x \ln x - \int \rd x + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds x \ln x - x + C\) | Primitive of Constant |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Integration: Integration by Parts: Example