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?

The easiest way is using undetermined coefficients:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (Ax+B)\sqrt{x^2+1}+\lambda\int{\frac{dx}{\sqrt{x^2+1}}}$$
You must differentiate and you have:
$$\frac{2x^2+3x+1}{\sqrt{x^2+1}}=A\sqrt{x^2+1}+(Ax+B)\frac{2x}{2\sqrt{x^2+1}}+\frac{\lambda}{\sqrt{x^2+1}}$$
So you have an equation:
$$2x^2+3x+1=A(x^2+1)+(Ax+B)x+\lambda$$
$$2x^2+3x+1=Ax^2+A+Ax^2+Bx+\lambda$$
from which you can find out an simultaneous equation:
\begin{cases}
2=2A\\
3=B\\
1=A+\lambda
\end{cases}
After solving it you have:
\begin{cases}
A=1\\B=3\\\lambda=0
\end{cases}
To remind, your integral grom the beginning is:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (Ax+B)\sqrt{x^2+1}+\lambda\int{\frac{dx}{\sqrt{x^2+1}}}$$
So 
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (x+3)\sqrt{x^2+1}$$
