Evaluate $$\int \frac{e^{2x}-1}{\sqrt{e^{2x}-(x+1)^2}} dx $$
Now, what I've checked is the derivative of $\sqrt{e^{2x}-(x+1)}$ = $\frac{2e^{2x}-1}{2\sqrt{e^{2x}-(x+1)}}$ which kind of gives me a way to start but it is not much . Any hints will be appreciated !
The problem is from a Romanian Mathematical Gazette from a older edition called in my language "Gazeta Matematica" ( I don't know if I wrote the name correctly in english) . It pretty much publishes problems for the olympiad.
$$I=\int \frac{e^{2x}-1}{\sqrt{e^{2x}-(1+x)^2}} dx$$ Let $$J=\int \frac{e^{2x}-(1+x)}{\sqrt{e^{2x}-(1+x)^2}}dx$$ Let $$e^{2x}-(1+x)^2=t \implies 2[e^{2x}-(1+x)] dx=dt$$ $$\implies J=\int \frac {dt}{2\sqrt{t}} dt =\sqrt{e^{2x}-(1+x)^2}$$ next note that $$I=J-K, ~~ K=\int \frac{x dx}{\sqrt{e^{2x}-(1+x)^2}}=\int \frac{x e^{-x}dx}{\sqrt{1-e^{-2x}(1+x)^2}}$$ Let $e^{-x}(1+x)=u \implies -xe^{-x} dx =du$, then $$K=-\int \frac{du}{\sqrt{1-u^2}}=-\sin^{-1}[e^{-x} (1+x)]$$ Finally, $$I=\sqrt{e^{2x}-(1+x)^2}+ \sin [e^{-x} (1+x)]+C$$