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?
 A: Try making a substitution of the inside function, that often works better. In this case, for the equation
 $$\int\sqrt{e^t-5}dt$$
make the substitution $u=e^t$, $\,\,du = e^tdt$
 $$=\int\frac{\sqrt{u-5}}{u}du$$
substitute $s = \sqrt{u-5}$, $\,\,ds = \frac{1}{2\sqrt{u-5}}$
 $$=2\int\frac{s^2}{s^2 + 5}ds$$
 $$=2\int\ \bigg(1-\frac{5}{s^2 + 5}\bigg)ds$$
 $$=2\int1\,ds - 10\int \frac{ds}{s^2 + 5}$$
 $$=2s - 2\int \frac{ds}{\frac{s^2}{5} + 1}$$
substitute $p = \frac{s}{\sqrt{5}}$, $\,\,dp = \frac{ds}{\sqrt{5}}$
 $$=2s - 2\sqrt{5}\int \frac{dp}{p^2 + 1}$$
 $$=2s - 2\sqrt{5}\tan^{-1}(p)$$
After a couple back-subtitutions we get
 $$=2\sqrt{e^t - 5} - 2\sqrt{5}\tan^{-1}\bigg(\frac{\sqrt{e^t - 5}}{\sqrt{5}}\bigg) + C$$
I like solving the problem this way as it feels more linear... Instead of trying to find what trick to use you just keep messing with the integral, making it simpler until you notice that you are either near to or in a form you can easily integrate.
A: Notice,  let $e^t-5=u^2\implies e^tdt=2udu$ $$dx=\frac{2udu}{u^2+5}$$
$$\int\sqrt{e^t-5}dt=\int u\frac{2udu}{u^2+5}$$
$$=2\int \frac{u^2du}{u^2+5}$$ $$=2\int \frac{u^2+5-5du}{u^2+5}$$
$$=2\int \frac{u^2+5du}{u^2+5}-10\int \frac{du}{u^2+5}$$
$$=2\int du-10\int \frac{du}{u^2+5}$$ $$=2u-10\frac{1}{\sqrt 5}\tan^{-1}\left(\frac{u}{\sqrt 5}\right)+C$$
$$=2\sqrt{e^t-5}-2\sqrt 5\tan^{-1}\left(\sqrt{\frac{e^t-5}{5}}\right)+C$$
A: Note that formally we have $u = \sqrt{e^{t}-5}$ iff $\log(u^{2}+5) = t$,
so
$$
\int_{t} \sqrt{e^{t}-5} = \int_{u = \sqrt{e^{t}-5}}u \cdot D\log(u^{2}+5) = \int_{u} \frac{2u^{2}}{u^{2}+5}. 
$$
Can you continue?
