I need help solving for this integral: $$\int \frac{e^x}{\sqrt{16-2^{2x}}}dx.$$
I tried using trigonometric substitution, but I wasn't able to solve it in that way ( I didn't get anything "friendly").
Thanks in advance.
2026-04-01 16:27:54.1775060874
Bumbble Comm
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Antiderivative of $\frac{e^x}{\sqrt{16-2^{2x}}}$
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Bumbble Comm
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Notice that $$2^{2x}=e^{\ln\left(2^{2x}\right)}=\left(e^x\right)^{\ln 4}$$ So: $$\int{\frac{e^x}{\sqrt{16-2^{2x}}}}dx=\int{\frac{e^x}{\sqrt{16-\left(e^x\right)^{\ln 4}}}}dx$$ Now:$$u=e^x,~du=e^x dx$$ So you get $$\int{\frac{1}{\sqrt{16-u^{\ln 4}}}}du$$ Now, unless you know VERY complicated things(the hypergeometric function) you can't do it, maybe you copied the function wrong?
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If you mean $$\int \frac{e^x}{\sqrt{16-e^{2x}}}dx$$ then $e^x=4\cos{t}$ helps.