Question

How do you integrate `x/(1+e^(2x^2))`?

Answer 1

We're going to need some substitutions, I'll make one at the time to be as clear as possible:

• First of all, you substitute `u=x^2`, to get `du=2xdx` and transform the integral into
  `1/2 \int{1}/{e^{2u}+1}du`
• Then, we'll go with `s=2u`. This means `ds=2du`, so we only have to add another factor `1/2`, and get
  `1/4 \int{1}/{e^{s}+1}ds`
• Now, let's get rid of the exponential, defining `p=e^s`. This means `dp=e^sds`. Multiplying and dividing by `e^s`, we get
  `1/4 \int{1}/{p(p+1)}dp`
• Writing `1/{p(p+1)}` as `1/p - 1/(p+1)`, we can split the integrals, and get
  `-1/4\int 1/{p+1} dp + 1/4\int 1/p dp`.
• These are both known integral, giving `log(p+1)` and `\log(p)`. Now we need to make all the substitutions backwards: from
  `-1/4\log(p+1)+ 1/4\log(p)`
• `p=e^s`, so we have
  `-1/4 \log(e^s+1) + 1/4\log(e^s)`
• `s=2u`, so we have
  `-1/4 \log(e^{2u}+1) + 1/4\log(e^{2u})`
• Finally, `u=x^2`, so we have
  `-1/4 \log(e^{2x^2}+1) + 1/4\log(e^{2x^2})`

The only further step we can do is writing the answer as

`-1/4 \log(e^{2x^2}+1) + x^2 /2`
since we have `\log(e^{2x^2})=2x^2`, and we divided by 4.