Intergral Partial Fraction Question

[markraz]

hi,

I have this integral

\(\displaystyle \int \frac{1}{2+e^x} dx\)

Apparently a "preliminary step" needs to be performed to convert to:

\(\displaystyle \int \frac{e^x}{2e^x+1}dx\)

What is the reason for the "preliminary step"?

 

[stapel]

I have this integral

\(\displaystyle \int \frac{1}{2+e^x} dx\)

Apparently a "preliminary step" needs to be performed to convert to:

\(\displaystyle \int \frac{e^x}{2e^x+1}dx\)

If the exponent on the "e" in the first integral is really -x, rather than just x, then the conversion is correct. Otherwise, it's obviously not correct.

What is the reason for the "preliminary step"?

At a guess, it makes u-substitution fairly simple.

 

[HallsofIvy]

If the problem was \(\displaystyle \int \frac{1}{2+ e^{-x}}dx\), then multiplying both numerator and denominator by \(\displaystyle e^x\) gives \(\displaystyle \int \frac{e^x}{2e^x+ 1}dx\). An "obvious" substitution would be \(\displaystyle u= 2e^x+ 1\). Do you see why you need the "\(\displaystyle e^x\)" in the numerator?

If the problem really was \(\displaystyle \int\frac{1}{2+ e^x}dx\), then multiplying both numerator and denominator by \(\displaystyle e^{-x}\) gives \(\displaystyle \int \frac{e^{-x}}{2e^{-x}+ 1}dx\) and an "obvious" substitution would be \(\displaystyle u= 2^{-x}+ 1\). Do you see why you need the "\(\displaystyle e^{-x}\)" in the numerator?