Primitive of Logarithm of x

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Theorem

$\ds \int \ln x \rd x = x \ln x - x + C$


Corollary

$\ds \int \map \ln {1 - x} \rd x = \paren {x - 1} \map \ln {1 - x} - x + C$


Proof 1

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

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

$\blacksquare$


Proof 2

Note that we have:

\(\ds \int_0^1 \ln x \rd x\) \(=\) \(\ds \int_0^1 x^0 \paren {\ln x}^1 \rd x\)
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^1 \map \Gamma 2} {1^2}\) Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$
\(\ds \) \(=\) \(\ds -1\) Gamma Function Extends Factorial

We therefore have:

\(\ds -1\) \(=\) \(\ds \int_0^1 \ln x \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 a \int_0^a \map \ln {\frac u a} \rd u\) substituting $x = \dfrac u a$ where $a > 0$
\(\ds \) \(=\) \(\ds \frac 1 a \int_0^a \paren {\ln u - \ln a} \rd u\) Difference of Logarithms
\(\ds \) \(=\) \(\ds \frac 1 a \int_0^a \ln u \rd u - \ln a\) Primitive of Constant, Fundamental Theorem of Calculus

giving:

$\ds \frac 1 a \int_0^a \ln u \rd u = \ln a - 1$

so:

$\ds \int_0^a \ln u \rd u = a \ln a - a$

for all real $a > 0$.

By Fundamental Theorem of Calculus: First Part, we have that:

$x \ln x - x$ is a primitive of $\ln x$ on $x > 0$.

We therefore conclude that, by Primitives which Differ by Constant:

$\ds \int \ln x \rd x = x \ln x - x + C$

for $x > 0$.

$\blacksquare$


Also presented as

Some sources present this result as:

$\ds \int \ln x \rd x = x \paren {\ln x - 1} + C$


Sources