How do you use substitution to integrate #(ln(5x)/x)*dx#?

Answer 1

The integral is evaluated below.

We have to evaluate the integral,

#int Ln (5x)/x*dx#
Let, #Ln (5x) = t# #implies Ln x + Ln 5 = t#
#therefore 1/x + 0 = (dt)/dx# (Differentiating with respect to #x#)
#implies dt = dx/x#
Now, let us attack our problem and substitute #dt# in place of #dx/x# and the integral becomes,
#int t*dt = t^2/2 + C#
In terms of #x#,
#int Ln (5x)/x*dx = (Ln (5x))^2/2 + C#
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Answer 2

To integrate the expression (\frac{{\ln(5x)}}{x} , dx) using substitution, let (u = \ln(5x)). Then (du = \frac{1}{x} dx).

Rewrite the integral in terms of (u): [ \int \frac{u}{5} , du ]

Now integrate with respect to (u): [ \frac{1}{10} u^2 + C ]

Substitute back for (u): [ \frac{1}{10} (\ln(5x))^2 + C ]

So, the integral of (\frac{{\ln(5x)}}{x} , dx) is (\frac{1}{10} (\ln(5x))^2 + C), where (C) is the constant of integration.

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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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