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