integral (e^(9x))/(e^(18x)-e^(9x))dx

Question

hartnn · Answer

did you try to put 
u = e^(9x) 
? 
du = ... ?

anonymous · Answer

i tried using substitution and then used partial fractions, but my answer was still wrong

dumbcow · Answer

$$\rightarrow \frac{1}{9} \int\limits \frac{du}{u^2 -u} =  \frac{1}{9} \int\limits \frac{du}{u(u-1)} =  \frac{1}{9} \int\limits \frac{du}{u -1} - \frac{1}{9} \int\limits \frac{du}{u}$$
$$= \frac{1}{9}\ln (u-1) -\frac{1}{9} \ln u $$
$$=\frac{1}{9} \ln (e^{9x} -1) - x$$

anonymous · Answer

i got the same answer, but the system read it as incorrect, so i think there might be something wrong with the system itself, it happens very often.

hartnn · Answer

maybe you forgot to add the "+c" in the end :P

anonymous · Answer

i tried to it with +constant and without so. i'll just email my teacher and confirm that i have my answer right.and get the point for that problem

hartnn · Answer

good luck!