How to Write an Answer for Integration with Logarithms?

[jiasyuen]

For example, $$\int \frac{e^x}{3e^x-1}dx$$,

Should I write my answer in this $$\frac{1}{3}\ln (3e^x-1)+c$$ or $$\frac{1}{3}\ln \left | 3e^x-1 \right |+c$$ ?

 

[chisigma]

For example, $$\int \frac{e^x}{3e^x-1}dx$$,

Should I write my answer in this $$\frac{1}{3}\ln (3e^x-1)+c$$ or $$\frac{1}{3}\ln \left | 3e^x-1 \right |+c$$ ?

$\displaystyle \int \frac{e^{x}}{3\ e^{x} - 1}\ d x = \frac{1}{3}\ \int \frac{e^{x}}{e^{x} - \frac{1}{3}}\ d x = \frac{1}{3}\ \ln |e^{x} - \frac{1}{3}| + c $

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

For example, $$\int \frac{e^x}{3e^x-1}dx$$,

Should I write my answer in this $$\frac{1}{3}\ln (3e^x-1)+c$$ or $$\frac{1}{3}\ln \left | 3e^x-1 \right |+c$$ ?

You can remove the absolute value symbols as long as:

$$3e^x-1>0$$

for all values of $x$ in the integrand's domain...what do you conclude?

 

[chisigma]

In my opinion it is possible to write...

$\displaystyle \int \frac{f^{\ '} (x)}{f(x)}\ d x = \ln f(x) + c\ (1)$

... no matter if in a given interval is $f(x) > 0$ or $f(x) < 0$ [but not $f(x)=0$...] , taking into account that for $f(x)<0$ is $\ln f(x) = \ln |f(x)| + i\ \pi$...

Kind regards

chi sigma