Integral of ln x

ln x = (ln x)⋅1

So solve this
by using integral by parts.

The order of uv' is

[u]
logarithmic (ln x)
polynomial (1)
trigonometric
exponential
[v'].

So set
u = ln x and v' = 1.

Write u = ln x.

Differentiate both sides.

Then u' = 1/x.

Derivative of ln x

Write v' = 1
next to u' = 1/x.

Integrate both sides.

Then v = x.

Integral of a Polynomial

Write this above v' = 1.

u = ln x, v = x
u' = 1/x

Then the given integral is equal to,
uv, (ln x)⋅x
minus
integral, u'v, (1/x)⋅x dx.

(ln x)⋅x = x ln x
(1/x)⋅x = 1

-∫ 1 dx = -x + C

So the integral of ln x is
x ln x - x + C.