Completing the square yields \(4x^2 + 8x -5= 4(x+1)^2 -
9\text{.}\) So, the substitution \(u=2(x+1)\) turns the integral into
\begin{equation*}
\frac{1}{2} \int \frac{1}{\sqrt{u^2-9}} du.
\end{equation*}
Note that the integrand \(f(u) = 1/\sqrt{u^2-9}\) is an even function of \(u\text{,}\) so if \(F(u)\) is an antiderivative of \(f(u)\) on \(u
\gt 3\text{,}\) then \(-F(-u)\) would be an anti-derivative \(f(u)\) on \(u \lt 3\text{.}\) On \(u \gt 3\text{,}\) we can evaluate the integral by the substitution \(u=3\sec(\theta)\text{:}\)
\begin{align*}
\frac{1}{2} \int \frac{3\sec(\theta)\tan(\theta)}{3\tan(\theta)} d\theta
& = \frac{1}{2}\int\sec(\theta) d\theta\\
& = \frac{1}{2}\ln|\sec(\theta)
+ \tan(\theta)|+C\\
& = \frac{1}{2}\ln\left| \frac{u}{3} + \frac{\sqrt{u^2-9}}{3}\right| + C\\
& = \frac{1}{2}\ln\left| u + \sqrt{u^2-9} \right| + C.
\end{align*}
Let \(F(u)\) be the anti-derivative of \(f(u)\) displayed above. Then \(\frac{F(u)-F(-u)}{2}\) will be an odd antiderivative of \(f(u)\) on the whole domain of \(f(u)\text{.}\) Note that
\begin{align*}
F(u)-F(-u) & =
\frac{1}{2}\left(\ln\left|u+\sqrt{u^2-9}\right| -
\ln\left|-u + \sqrt{(-u)^2-9}\right|\right) \\
&= \frac{1}{2}\ln
\left|\frac{u+\sqrt{u^2-9}}{u-\sqrt{u^2-9}}\right| \\
& = \frac{1}{2}\ln\left|\frac{(u+\sqrt{u^2-9})^2}{9}\right|\\
&= \ln|u+\sqrt{u^2-9}|-\ln(3).
\end{align*}
Therefore, the integral can be expressed as
\begin{equation*}
\frac{1}{2}\ln|u+\sqrt{u^2-9}|+C = \frac{1}{2}\ln|2(x+1) +
\sqrt{4x^2+8x-5}| +C
\end{equation*}
where \(C\) denotes the class of locally constant functions on the domain of the integrand.
