Question

Let $I=\int \frac{e^x}{e^{4x}+1}dx$ and $J=\int \frac{e^{-x}}{e^{-4x}+1}dx$
Then for any arbitrary constant C, match the following

Answer 1

$I=\int \frac{e^x}{e^{4x}+1}dx$
Substituting $e^x=t\Rightarrow e^{x}dx=dt$, we get
$I=\int \frac{1}{t^4+1}dt=\int (\frac{\sqrt{2}t-2}{4(-t^2+\sqrt{2}t-1)}+\frac{\sqrt{2}t+2}{4(t^2+\sqrt{2}t+1)})dt=\frac{1}{4}\int \frac{\sqrt{2}t+2}{(t^2+\sqrt{2}t+1)}dt+\frac{1}{4}\int \frac{\sqrt{2}t-2}{(-t^2+\sqrt{2}t-1)}dt=\frac{1}{4}\int (\frac{\sqrt{2}t+2}{\sqrt{2}(t^2+\sqrt{2}t+1)}+\frac{1}{(t^2+\sqrt{2}t+1)})dt+\frac{1}{4}\int (\frac{-\sqrt{2}t-2}{\sqrt{2}(-t^2+\sqrt{2}t-1)}-\frac{1}{(-t^2+\sqrt{2}t-1)})dt=-\frac{\log (-t^2+\sqrt{2}t-1)}{2\sqrt{2}}+\frac{\log (t^2+\sqrt{2}t+1)}{4\sqrt{2}}-\frac{ta{n}^{-1}(1-\sqrt{2}t)}{2\sqrt{2}}+\frac{ta{n}^{-1}(1+\sqrt{2}t)}{2\sqrt{2}}+c=\frac{1}{2\sqrt{2}}(ta{n}^{-1}\left(\frac{e^{2x}-1}{\sqrt{2}{e}^x}\right)-\frac{1}{2}\ln \left(\frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}\right))+c$
$J=\int \frac{e^{-x}}{e^{-4x}+1}dx$
Substituting $e^{-x}=t\Rightarrow -e^{-x}dx=dt$, we get
$J=-\int \frac{1}{t^4+1}dt=\frac{1}{2\sqrt{2}}(-ta{n}^{-1}\left(\frac{e^{-2x}-1}{\sqrt{2}{e}^x}\right)+\frac{1}{2}\ln \left(\frac{e^{-2x}-\sqrt{2}{e}^{-x}+1}{e^{-2x}+\sqrt{2}{e}^{-x}+1}\right))+c=\frac{1}{2\sqrt{2}}(ta{n}^{-1}\left(\frac{e^{2x}-1}{\sqrt{2}{e}^x}\right)-\frac{1}{2}\ln \left(\frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}\right))+cJ+I=\frac{1}{\sqrt{2}}\left(ta{n}^{-1}\left(\frac{e^{2x}-1}{\sqrt{2}{e}^x}\right)\right)+cJ-I=\frac{1}{2\sqrt{2}}\ln \left(\frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}\right)+c$