integral of e^x+1 / e^x -1 ?

Question

solomonzelman · Answer

$$\LARGE\color{blue}{ \int\limits_{ }^{ } \frac{e^x+1}{e^x-1}  }$$ expand it to
$$\LARGE\color{blue}{ \int\limits_{ }^{ } \frac{e^x}{e^x-1}~~dx+\int\limits_{ }^{ } \frac{1}{e^x-1}~~dx  }$$

anonymous · Answer

i did that that thing....but im stucked in the 2nd equation

solomonzelman · Answer

$$\LARGE\color{blue}{ \int\limits_{ }^{ } \frac{1}{u}~~du+\int\limits_{ }^{ } \frac{1}{e^x-1}~~dx  }$$

solomonzelman · Answer

Tell me what you get for the first integral

anonymous · Answer

|dw:1407250080475:dw|

anonymous · Answer

im thinking of multiplying it with$$\frac{ 1 }{ e ^{x}-1 }\frac{ e ^{x}+1 }{ e ^{x}+1 }$$

anonymous · Answer

i would go with partial fractions

solomonzelman · Answer

Tell me what you get for the first integral

anonymous · Answer

$$\frac{e^x+1}{e^x-1}=\frac{e^x-1+2}{e^x-1}=1+\frac{2}{e^x-1}$$ first is obvious, second use $u=e^x-1$

solomonzelman · Answer

yes and then sub  s  for e^x  (I think)

anonymous · Answer

i would go the whole hog with the $u$ sub 
$$u=e^x-1\
u+1=e^x\
\log(u+1)=x$$ making 
$$dx=\frac{1}{u+1}$$ and the integral would then be 
$$2\int \frac{du}{u(u+1)}$$

anonymous · Answer

i got it now tnx