<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$$\int (x+1)\sqrt{x^2+1}dx=$ $\int\frac{t^2-2t-1}{2t} \frac{1-t^2}{2t} \frac{t^2+1}{2t}dt=$ 
$\frac{1}{8} \int \frac{-t^6+2t^5+t^4+t^2-2t-1}{t^3}dt=$
$\frac{1}{8} [-\int t^3 dt+2\int t^2 dt+\int t dt +\int \frac{1}{t} dt-2 \int \frac{1}{t^2} dt- \int \frac{1}{t^3}dt]=$ 
$\frac{1}{8}[-\frac{t^4}{4}+2\frac{t^3}{3}+\frac{t^2}{2}+lnt+\frac{2}{t}-\frac{1}{2t^2}]$
We do the substitution from $(2)$ and we get the result
You think that's correct?
 [/INST] After the rationalizing substitution, I think you made a few errors. It should be
$$\int (x+1)\sqrt{x^2+1}\,dx=\int\frac{t^2-2t-1}{2t}\cdot \frac{t^2+1}{2t}\cdot \frac{t^2+1}{2t^2}dt.$$
Alternative approach by using hyperbolic functions. 
Let $x=\sinh(t)$ then $\cosh^2(t)-\sinh^2(t)=1$ and
$$\begin{align*}\int (x+1)\sqrt{x^2+1}\,dx &=\int (\sinh(t)+1)\sqrt{\sinh^2(t)+1}\,\cosh(t) dt \\
&=\int (\sinh(t)+1)\cosh^2(t) dt\\
&=\frac{\cosh^3(t)}{3}+ \frac{\cosh(t)\sinh(t)}{2}+\frac{t}{2}+c\\
&=\frac{(x^2+1)^{3/2}}{3}+\frac{x(x^2+1)^{1/2}}{2}+\frac{\text{arcsinh}(x)}{2}+c
\end{align*}.$$
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