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

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

Answer

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.

So the final answer is:

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Therefore,

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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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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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