Difficult Integral I have a problem with solving this integral:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx$$
I tried to use substitution but I got stuck. Can anyone help me?

You can also use one of the Euler's substitutions:
$$\begin{align}
  \sqrt{x^2+1} &= t-x \\ 
   x &= \frac{1}{2}\left(t - \frac{1}{t}\right) \\
   dx &= \frac{1}{2}\left(1 + \frac{1}{t^2}\right)dt \\
   \sqrt{x^2+1} &= \frac{1}{2}\left(t + \frac{1}{t}\right) \\
   \frac{dx}{\sqrt{x^2+1}}&=\frac{dt}{t}\\
   2x^2+3x+1&= \frac{t^4+3t^3-3t+1}{2t^2}
  \end{align}
$$
Integration is now straightforward
$$\begin{align}\int\frac{t^4+3t^3-3t+1}{2t^3}dt&=\frac{-t + 6t + 6 t^3 + t^4}{4 t^2}+C\\&=\frac{(t^2+1)(t^2+6 t-1)}{4 t^2}+C \\
&=\frac{1}{2}\left(1+\frac{1}{t}\right)\left[\frac{1}{2}\left(t-\frac{1}{t}\right)+3\right]\\
&=\sqrt{x^2+1}(x+3)
\end{align}$$
As it often happens with the Euler's method, the algebra is tedious, but it gets you there.
