<s>[INST] Integral by the given $u$-substitution Original equation $$\int\sqrt{e^t-5}dt \qquad u=\sqrt{e^t-5}$$
This is what I've done so far$$\int{u}dt$$ $$du=\frac{e^t}{2\sqrt{e^t-5}}dt$$$$dt=\frac{2\sqrt{e^t-5}}{e^t}du$$$$\int{u}{\frac{2\sqrt{e^t-5}}{e^t}}du$$ From this point on what would i need to do?
 [/INST] Notice,  let $e^t-5=u^2\implies e^tdt=2udu$ $$dx=\frac{2udu}{u^2+5}$$
$$\int\sqrt{e^t-5}dt=\int u\frac{2udu}{u^2+5}$$
$$=2\int \frac{u^2du}{u^2+5}$$ $$=2\int \frac{u^2+5-5du}{u^2+5}$$
$$=2\int \frac{u^2+5du}{u^2+5}-10\int \frac{du}{u^2+5}$$
$$=2\int du-10\int \frac{du}{u^2+5}$$ $$=2u-10\frac{1}{\sqrt 5}\tan^{-1}\left(\frac{u}{\sqrt 5}\right)+C$$
$$=2\sqrt{e^t-5}-2\sqrt 5\tan^{-1}\left(\sqrt{\frac{e^t-5}{5}}\right)+C$$
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