<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$$\int (x+1)\sqrt{x^2+1}dx=$ $\int\frac{t^2-2t-1}{2t} \frac{1-t^2}{2t} \frac{t^2+1}{2t}dt=$ 
$\frac{1}{8} \int \frac{-t^6+2t^5+t^4+t^2-2t-1}{t^3}dt=$
$\frac{1}{8} [-\int t^3 dt+2\int t^2 dt+\int t dt +\int \frac{1}{t} dt-2 \int \frac{1}{t^2} dt- \int \frac{1}{t^3}dt]=$ 
$\frac{1}{8}[-\frac{t^4}{4}+2\frac{t^3}{3}+\frac{t^2}{2}+lnt+\frac{2}{t}-\frac{1}{2t^2}]$
We do the substitution from $(2)$ and we get the result
You think that's correct?
 [/INST] Before trying the substitution, it is worth to simplify the integrand, by observing that
$$(x^2+1)'=2x$$ so that
$$\int(x+1)\sqrt{x^2+1}\,dx=\frac13(x^2+1)^{3/2}+\int\sqrt{x^2+1}\,dx.$$
For the last integral, substitution with $x=\sinh t$ will allow to get rid of the square root. By parts intégration also works:
$$I:=\int\sqrt{x^2+1}\,dx=x\sqrt{x^2+1}-\int\frac{x^2}{\sqrt{x^2+1}}dx$$
and
$$\int\frac{x^2}{\sqrt{x^2+1}}dx=\int\frac{x^2+1-1}{\sqrt{x^2+1}}dx=I-\text{arshinh } x,$$ from which you draw $I$.
</s>