In this opportunity I want to present the natural logarithm theme, for that we draw the graph of the function f (t) = 1 / t, where t ≠ 0, and we focus our attention on the first quadrant of the system of coordinate axes.
Let's see the image:

Let's focus our attention on the area R under that curve, limited by the lines x = 1 and x=t, where x> 1 ( x can also be 0 <x <1)
In the following image the area R is shaded:

That area R limited by the lines x = 1, x=t where x can be greater or less than 1 (x is always greater than 0) and the curve y = 1 / t (where it must be different from 0); receive the name of the Neperian Logarithm of x, and denote Lnx.
That is to say that the Neperian logarithm is an area between curves, so it is expressed as an integral, see formula in the image.

Applying the Rule of the Chain, we see that:

Let's see an application example:
D x Ln (5x⁴ -1) = [1 / (5x⁴ -1)] D x [(5x⁴ -1] = [1 / (5x² -1)] [20x] = 20x / (5x 4 -1)
You can also prove that:
D x Ln | x | = 1 / x where x ≠ 0.

Let's suppose x> 0, so | x | = x, and D x Ln | x | = D x Lnx = (1 / x). (1) = 1 / x
If x <0, so D x Ln | x | = D x Ln (-x) = (-1 / x) (- 1) = 1 / x
Thus:
∫1 / xdx = Ln | x | + C, donde x ≠ 0

Generalizing through the "Rule of the chain"
It is concluded that:
∫ (1 / u) du = Ln | u | + C, donde u ≠ 0

Let's see the following application example.
∫ [x / (4x² +1)] dx

We do u = 4x² +1
So du = 8xdx
Substituting in the integral, we have:

∫ [x / (4x² +1)] dx = (1 / 8)∫du / u = (1 / 8)∫ [1 / u] du = 1/8 [ln | u | + c] = 1/8 [ln | 4x² +1 | + c]
= (1/8) ln | 4x² +1 | + (1/8) c = (1/8) ln | 4x² +1 | + k; where the integration constant k = (1/8)c

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Reference:
https://docs.google.com/file/d/0B8wx9bEA31Jzc0tyTVlWOHgycnc/edit
The images are of my own creation.