Question

How do you integrate `ln(5x+3)`?

Answer 1

Edit: I misread the question- I didn't integrate, I differentiated.

Answer

`d/dx ln(5x+2)= 5/(5x+3)`

Solution

You would do this using the chain rule.

The chain rule, in words, basically just means:

the derivative of the outer function(leaving the inner function alone, or treating it as a single variable) X the derivative of the inner function

Here, seeing the "outer" and "inner" functions is pretty straightforward.

We have `ln(5x+3)`. Just by looking at it, you can see that `5x+3` is "inside" the `ln`, making it the inner function.

Now we can do the chain rule. We know that the derivative of `ln (u)`, for example, is just `1/u`. Well, the derivative of `ln(5x+3)` (while leaving the inner function alone, or treating is as "u"!) is `1/(5x+3)`. But now, to complete the chain rule, we have to multiply by the derivative of the inner function- The derivative of `5x+3` is simply `5`.

So the final answer is:

`d/dx ln(5x+2)= 1/(5x+3) * 5 = 5/(5x+3)`

Answer 2

`int ln(5x+3) dx`.

Let `w=5x+3`, so that `dw = 5dx` and the integral becomes:

`1/5 int lnw dw`

Integrate by parts: `u = lnw` and `dv = dw` this makes:

`du = 1/w dw` and `v = w`, using the formula for integral by parts:

`1/5 int lnw dw = 1/5[w lnw - int w*1/w dw]`

`color(white)"ssssssssssss"` `= 1/5 [wlnw - int dw]`

`color(white)"ssssssssssss"` `= 1/5 [wlnw - w]+C`

Therefore,

`int ln(5x+3) dx = 1/5 [(5x+3)ln(5x+3) - (5x+3)]+C`.

`color(white)"sssssssssssssss"` `=(5x+3)/5 (ln(5x+3) - 1)+C`