<s>[INST] Integration by substitution to take out square root 
Find $$\int (x+1)\sqrt{x^2+1}\,dx .$$

In order to not bother with the square root I thought of doing this:
$let$ $ x^2+1=(x+t)^2$ $\Rightarrow$ $x=\frac{1-t^2}{2t}$ $\Rightarrow x-t=\frac{1+t^2}{2t}$ $(1)$
$\Rightarrow$ $t=\sqrt{x^2+1}-x$ $(2)$
And
$dx=-\frac{t^2+1}{2t^2}dt $ $(3)$
and also $x+1=-\frac{t^2-2t-1}{2t}$ $(4)$
Then
$(1),(3),(4)\Rightarrow$$\int (x+1)\sqrt{x^2+1}dx=$ $\int\frac{t^2-2t-1}{2t} \frac{1-t^2}{2t} \frac{t^2+1}{2t}dt=$ 
$\frac{1}{8} \int \frac{-t^6+2t^5+t^4+t^2-2t-1}{t^3}dt=$
$\frac{1}{8} [-\int t^3 dt+2\int t^2 dt+\int t dt +\int \frac{1}{t} dt-2 \int \frac{1}{t^2} dt- \int \frac{1}{t^3}dt]=$ 
$\frac{1}{8}[-\frac{t^4}{4}+2\frac{t^3}{3}+\frac{t^2}{2}+lnt+\frac{2}{t}-\frac{1}{2t^2}]$
We do the substitution from $(2)$ and we get the result
You think that's correct?
 [/INST] Avoiding hyperbolic functions, I believe you can do a trigonometric substitution: consider a right triangle with angle $\theta$ for which the opposite side is $x$, the adjacent side is $1$, and (thus) the hypotenuse is $\sqrt{x^2 + 1}$.
Then $x = \tan{\theta}$, so $dx = \sec^2{\theta}d\theta$; note that $\sec{\theta} = \sqrt{x^2 + 1}$ and your integral becomes:
$$\int(x+1)\sqrt{x^2 + 1}dx = \int(\tan{\theta} + 1)\sec^3{\theta} d\theta = \int\tan{\theta}\sec^3{\theta} + \sec^3{\theta} d \theta$$
The first part of the integrand is not difficult, since $u = \sec{\theta}$ yields $du = \sec{\theta}\tan{\theta}d\theta$, whereby
$$\int \tan\theta \sec^3\theta d\theta = \int u^2 du$$
and finding the antiderivative of secant cubed is now the only task that remains. If you have seen this before (example link) then you have reduced the problem to something familiar. If not, then you can use the already provided answer(s) for an alternative derivation of this function's antiderivative!
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