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?
 A: Here is another way of doing this integral:
Let $x=\sinh u\Rightarrow dx=\cosh du$. Also $\sqrt{x^2+1}=\cosh u$.
Then $$I=\int (2\sinh^2u+2\sinh u+1 )du$$
$$=\int(\cosh 2u+3\sinh u) du$$
$$=\frac 12 \sinh2u+3\cosh u +c$$
$$=(\sinh u+3)\cosh u+c$$
$$=(x+3)\sqrt{x^2+1}+c$$
A: 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 
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (x+3)\sqrt{x^2+1}$$
A: Let $y=\arctan x,x=\tan y\implies\sec y=+\sqrt{x^2+1}$ as $-\dfrac\pi2\le y\le\dfrac\pi2$
$$\int\dfrac{2x^2+3x+1}{\sqrt{x^2+1}}dx=\int(2\tan^2y+3\tan y+1)\sec y\ dy$$
$$=2\int\sec^3y\ dy-\int\sec y\ dy+3\int\tan y\ dy$$
Utilize  Indefinite integral of secant cubed   and replace the values of $\sec y,\tan y$
A: $$\int\frac{2x^2+3x+1}{\sqrt{x^2+1}}\space\text{d}x=$$

For the integrand $\frac{2x^2+3x+1}{\sqrt{x^2+1}}$, substitute $x=\tan(u)$ and $\text{d}x=\sec(u)\space\text{d}u$. 
Then $\sqrt{x^2+1}=\sqrt{\tan^2(u)+1}=\sec(u)$ and $u=\arctan(u)$:

$$\int\left(2\tan^2(u)+3\tan(u)+1\right)\sec(u)\space\text{d}u=$$
$$\int\left(\sec(u)+2\tan^2(u)\sec(u)+3\tan(u)\sec(u)\right)\space\text{d}u=$$
$$\int\sec(u)\space\text{d}u+\int 2\tan^2(u)\sec(u)\space\text{d}u+\int3\tan(u)\sec(u)\space\text{d}u=$$
$$\int\sec(u)\space\text{d}u+2\int \tan^2(u)\sec(u)\space\text{d}u+3\int\tan(u)\sec(u)\space\text{d}u=$$
$$2\int\sec^3(u)\space\text{d}u-\int\sec(u)\space\text{d}u+3\int\tan(u)\sec(u)\space\text{d}u=$$

Use the reduction formula:

$$\tan(u)\sec(u)+3\int\tan(u)\sec(u)\space\text{d}u=$$

Substitute $s=\sec(u)$ and $\text{d}s=\tan(u)\sec(u)\space\text{d}u$:

$$\tan(u)\sec(u)+3\int1\space\text{d}s=$$
$$\tan(u)\sec(u)+3s+\text{C}=$$
$$\tan(u)\sec(u)+3\sec(u)+\text{C}=$$
$$\tan\left(\arctan(x)\right)\sec\left(\arctan(x)\right)+3\sec\left(\arctan(x)\right)+\text{C}=$$
$$(x+3)\sqrt{x^2+1}+\text{C}$$
A: 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.
A: HINT:
Let $$\dfrac{2x^2+3x+1}{\sqrt{x^2+1}}=\dfrac{2(x^2+1)+3x-1}{\sqrt{x^2+1}}=2\sqrt{x^2+1}+\dfrac32\cdot\dfrac{2x}{\sqrt{x^2+1}}-\dfrac1{\sqrt{x^2+1}}$$
Use this
A: $$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx=\int{\frac{2\sinh^2(t)+3\sinh(t)+1}{\sqrt{\sinh^2(x)+1}}} \cosh(t)dt=
\int(\cosh(2t)+3\sinh(t))dt=\frac12\sinh(2t)+3\cosh(t)=x\sqrt{x^2+1}+3\sqrt{x^2+1}$$
