MHB 206.08.04.59 int completing the square

[karush]

$\tiny{206.08.04.59}$
$\textrm{Solve by completing the square}$

\begin{align*}\displaystyle
I_{41}&=\int \frac{1}{\sqrt[]{x^2+2x+37}} \, dx\\
\end{align*}
$\textit{from the radical we have}$
\begin{align*}\displaystyle
x^2+2x+37&=x^2+2x+1 +37-1\\
&=(x+1)^2 + 36
\end{align*}
$\textit{U substitution we have}$
\begin{align*}\displaystyle
u=x+1 \therefore du=dx\\
\end{align*}
$\textit{Thus the Integral now is:}$
\begin{align*}\displaystyle
&=\int \frac{1}{\sqrt{u^2 + 6^2}} \, du\\
\end{align*}
$\textit{then $a=6$ so from}$
\begin{align*}\displaystyle
\int\frac{1}{\sqrt{u^2 + a^2}}du&=\ln{|u+\sqrt{u^2+a^2}|}
\end{align*}
$\textit{finally}$
\begin{align*}\displaystyle
I_{41}&=\ln{\left|(x+1)+\sqrt{(x+1)^2+36}\right|}\\
\end{align*}

I hope anyway

 

[MarkFL]

The only thing I see missing is the constant of integration. Note that by using a hyperbolic trig. substitution, you could also write:

$$I=\arsinh\left(\frac{x+1}{6}\right)+C$$

 

[karush]

yeah
that would be better!