Primitive of Logarithm of x

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:

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

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

$\blacksquare$

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:

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