Why is there no integral of ln or log?

[Jason76]

No answer to $\displaystyle \int \log (x) dx$ or $\displaystyle \int \ln (x) dx$

However,$\displaystyle y' = \ln x = \dfrac{1}{x}$ and the $\displaystyle \int \dfrac{1}{x} dx = \ln (x) + C$ OR

However,$\displaystyle y' = \log x = \dfrac{1}{x}$ and the $\displaystyle \int \dfrac{1}{x} dx = \log (x) + C$

 

[Deleted member 4993]

No answer to $\displaystyle \int \log (x) or \int \ln (x)$

because $\displaystyle \int \ln (x)$ does not make sense.

However:

$\displaystyle \int \ln (x) \ \ dx = x * ln(x) - x + C$

 

[Jason76]

because $\displaystyle \int \ln (x)$ does not make sense.

However:

$\displaystyle \int \ln (x) \ \ dx = x * ln(x) - x + C$

Ok, edited original post, but what $\displaystyle \int \log (x) dx$ ? I know the other one was solved by using "integration by parts" or $\displaystyle \int uv' dx = uv - \int vu' dx$ where

$\displaystyle u = \ln x$

$\displaystyle v' = 1$

$\displaystyle v = x$

$\displaystyle u' = \dfrac{1}{x}$

 

[Deleted member 4993]

Ok, edited original post, but what $\displaystyle \int \log (x) dx$ ? I know the other one was solved by using "integration by parts" or $\displaystyle \int uv' = uv - \int vu'$ where

$\displaystyle u = \ln x$

$\displaystyle v' = 1$

$\displaystyle v = x$

$\displaystyle u' = \dfrac{1}{x}$

log₁₀x = log₁₀e * log_ex ............................ you need to brush up your algebra.

 

[lookagain]

No answer to $\displaystyle \int \log (x) dx$ or $\displaystyle \int \ln (x) dx$

However,$\displaystyle y' = \ln x = \dfrac{1}{x}$ No, I already stated elsewhere that the function is not equal to its derivative. and the \(\displaystyle \int \dfrac{1}{x} dx = \ln > > (x) < < + C\) Absolute value bars are needed around the "x" here.[/tex]

.

I know the other one was solved by using "integration by parts" or $\displaystyle \int uv' dx = uv - \int vu' dx$ where

The integration by parts rule is this:

$\displaystyle \int u \ dv \ = \ uv \ - \ \int v \ du$

$\displaystyle Here, \ \ \ u = ln(x), \ \ dv = 1 \ dx, \ \ du = (\frac{1}{x})dx, \ \ and \ \ v = x.$