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

2 Answers
May 19, 2015

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)#

May 19, 2015

#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#