Hi I am trying to find the integral $\int \frac{x}{e^{x^2} +1}\,dx$. The only techniques I know of so far are u-sub, integration by parts, and inspection. Can this be solved by these methods?

Like when I stare at this I don't see integration by parts helping because it wouldn't make it simpler by choosing that e value as either a u or dv. I was only able to get so far with u-sub till I got stuck.

  • $\begingroup$ try using the u-sub : $u = x^2$ $\endgroup$ – rapidracim Jun 15 '17 at 20:56
  • $\begingroup$ We don't usually integrate equations, but functions or differential forms. $\endgroup$ – Jack D'Aurizio Jun 16 '17 at 8:45

Substitute $u=x^2$.

Then $du = 2x \,dx$.

Therefore, $\displaystyle \int \frac{x}{e^{x^2}+1}\,dx=\frac{1}{2}\int\frac{1}{e^{(x^2)}+1}(2x\,dx)=\frac{1}{2}\int \frac{1}{e^u+1}\,du=\frac{1}{2}\int \left(1-\frac{e^u}{1+e^u}\right)\,du=$

$\displaystyle \frac{1}{2}(u-\ln(1+e^u)+C)=\boxed{\frac{1}{2}x^2-\frac{1}{2}\ln\left(1+e^{x^2}\right)+C}$

  • $\begingroup$ Missing factor $2$. $\endgroup$ – Yves Daoust Jun 15 '17 at 22:27
  • $\begingroup$ Thank you. Edited @YvesDaoust $\endgroup$ – Saketh Malyala Jun 15 '17 at 22:32
  • $\begingroup$ wait, one second $\endgroup$ – Saketh Malyala Jun 15 '17 at 22:33
  • $\begingroup$ Hem, applied the wrong way... $\endgroup$ – Yves Daoust Jun 15 '17 at 22:33
  • 1
    $\begingroup$ ALRIGHT HAHA got it this time! @YvesDaoust $\endgroup$ – Saketh Malyala Jun 15 '17 at 22:35


$$\frac x{e^{x^2}+1}=\frac{xe^{-x^2}}{1+e^{-x^2}}.$$

A substitution is now possible.

  • 2
    $\begingroup$ All too easy (+1) $\endgroup$ – Mark Viola Jun 15 '17 at 20:58


Use the substitution $$ u=e^{x^2}+1 \quad \rightarrow \quad xdx=\frac{1}{2}\frac{du}{e^{x^2}}=\frac{1}{2}\frac{du}{u-1 } $$

than use partial fraction decomposition.


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