Integrating Ln(x) with Two Easy Methods

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In summary, there are two easy ways to integrate ln(x): partial integration and differentiation with respect to a parameter. In method 1, we can use Leibnitz's rule to rewrite the integral of ln(x) as d[x ln(x)] - dx. In method 2, we can differentiate x^p with respect to the parameter p, set p equal to zero, and integrate the resulting function.
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matt_crouch
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how do you intergrate lnx?
 
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There are two easy ways:

1) Partial integration

2) Differentiation w.r.t. a parameter.

Method 1):

Leibnitz's rule can be written as:

d(fg) = f dg + g df

We can rewrite this as:

f dg = d(fg) - g df (1)

So, if we want to integrate ln(x), we can write according to Eq. (1):

ln(x) dx = d[x ln(x)] - x d[ln(x)] = d[xln(x)] - dx

So, the integral of ln(x) dx is the integral of d[xln(x)] minus the integral of dx, which is x ln(x) - x plus an arbitrary constant.


Method 2):

Consider integrating the function x^p. Then differentiate both sides w.r.t. the parameter p. Then set p equal to zero. Try it!
 

FAQ: Integrating Ln(x) with Two Easy Methods

How do you integrate lnx using the power rule?

The power rule states that the integral of x^n is (x^(n+1))/(n+1) + C. Therefore, to integrate lnx, we can rewrite it as x^1 and use the power rule to get the integral of lnx as (x^2)/2 + C.

Is there a specific method for integrating lnx?

Yes, there are various methods for integrating lnx, including the power rule, substitution, and integration by parts. The best method to use depends on the specific problem and its complexity.

What is the difference between integrating lnx and ln|x|?

The difference between integrating lnx and ln|x| lies in the domain of the function. While lnx is only defined for positive values of x, ln|x| is defined for both positive and negative values of x. This means that the integral of lnx will only give a positive result, while the integral of ln|x| can be both positive and negative.

Can lnx be integrated using u-substitution?

Yes, lnx can be integrated using u-substitution by letting u = lnx and du = (1/x)dx. This will transform the integral into ∫u du, which can be easily evaluated as u^2/2 + C, and then substituting back in for u to get the final answer of (lnx)^2/2 + C.

Can lnx be integrated using integration by parts?

Yes, lnx can be integrated using integration by parts by letting u = lnx and dv = dx. This will give du = (1/x)dx and v = x. Applying the integration by parts formula, we get the integral of lnx as xlnx - ∫1dx, which simplifies to xlnx - x + C.

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