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)#
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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#
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Answer 3

To integrate ( \ln(5x+3) ), you can use integration by parts:

[ \int \ln(5x+3) , dx = x \ln(5x+3) - \int \frac{x}{5x+3} , dx ]

Then, you can solve the integral ( \int \frac{x}{5x+3} , dx ) using u-substitution:

Let ( u = 5x+3 ), then ( du = 5 , dx ) or ( dx = \frac{1}{5} , du ).

Substitute ( u ) and ( dx ) back into the integral:

[ \int \frac{x}{5x+3} , dx = \frac{1}{5} \int \frac{1}{u} , du = \frac{1}{5} \ln|u| + C ]

[ = \frac{1}{5} \ln|5x+3| + C ]

Substitute this result back into the integration by parts formula:

[ \int \ln(5x+3) , dx = x \ln(5x+3) - \frac{1}{5} \ln|5x+3| + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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