Primitive of Logarithm of x/Proof 2

Proof
Note that we have:

We therefore have:

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 $x > 0$.