Integral of ln

[Jason76]

What is the integral of $\displaystyle \ln$? Example $\displaystyle \int \ln(3x + 5) = ?$ I know that the integral of a 1 over an expression is a $\displaystyle \ln$ expression. For instance
$\displaystyle \int \dfrac{1}{3x + 5} dx = \ln(3x + 5) + C$

 

[daon2]

I believe you are in a first Calculus course. The technique for integrating a logarithm is not taught until (in my experience) a second course in calculus. The technique is called integration by parts. See: http://en.wikipedia.org/wiki/Integration_by_parts

The answer is $\displaystyle \int \ln(x) dx = x\ln(x)-x + C = x(\ln(x)-1) + C$

 

[Jason76]

I believe you are in a first Calculus course. The technique for integrating a logarithm is not taught until (in my experience) a second course in calculus. The technique is called integration by parts. See: http://en.wikipedia.org/wiki/Integration_by_parts

The answer is $\displaystyle \int \ln(x) dx = x\ln(x)-x + C = x(\ln(x)-1) + C$

Thanks, so that's the general formula? What about $\displaystyle \int \ln(4x - 3) dx$?

 

[MarkFL]

In order to apply the formula give by daon2, you will want to use a u-substitution first:

$\displaystyle u=4x-3\,\therefore\,du=4\,dx$

and then your integral becomes:

$\displaystyle \displaystyle \frac{1}{4}\int\ln(u)\,du$

Now you can apply the formula, then back-substitute for u.

 

[Deleted member 4993]

What is the integral of $\displaystyle \ln$? Example $\displaystyle \int \ln(3x + 5) = ?$ I know that the integral of a 1 over an expression is a $\displaystyle \ln$ expression. For instance
$\displaystyle \int \dfrac{1}{3x + 5} dx = \ln(3x + 5) + C$.......... Incorrect

$\displaystyle \int \dfrac{1}{3x + 5} dx = \frac{1}{3}\ln(|3x + 5|) + C$

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