The logic used by the current solution is not correct. Moreover, it is not true that the general anti-derivative of $f(x)=e^{g(x)}$ is ${\frac {1}{g'(x)}}e^{g(x)}+c$ (take, for example, $g(x)=x^{2}$).

Nicholas Hu (talk)

Yeah whoops! Let me fix that.

Chong