Integration of $\frac{dx}{x\sqrt{\ln x^2 +6\ln x}}$

Question

$\require{cancel}$ Hello everybody so this is my integration question. I have reached a point in my picture. I don't know how to continue. I appreciate the help. Also , am I going in the right direction ??

Please note, square root is over $\ln x^2 +6\ln x$.

\begin{align*} \int\frac{1}{x\sqrt{\ln x^2+6\ln x}}dx\underset{\substack{\big| \\ u=\ln t \\ du=\frac{dt}{t}}}{=}{}&\int\frac{1}{\sqrt{u^2+6u}}du\underset{\substack{\big| \\ \text{Complete} \\ \text{square.}}}{=}\int\frac{1}{\sqrt{(u+5)^2-9}}du={} \\ {}\underset{\substack{\big| \\ s=u+5 \\ ds=du}}{=}{}&\int\frac{1}{\sqrt{s^2-9}}ds\underset{\substack{\big| \\ s=3\sec z \\ ds=3\tan z\sec zdz}}{=}\int\frac{1}{\sqrt{9\sec^2z-9}}3\tan z\sec zdz={} \\ {}\underset{\substack{\big| \\ \sec^2z-1=\tan^2z}}{=}{}&\int\frac{1}{\sqrt{9\tan^2z}}dz=\int\frac{3\tan z\sec z}{3\tan z}dz=\int\sec zdz={} \\ {}={}&\int\frac{\sec z(\tan z+\sec z)}{\tan z+\sec z}dz=\int\frac{\cancel{\sec z\tan z}+\sec^2z}{\cancel{\sec z\tan z}}dz=\int\sec^2zdz. \end{align*}

Image of work by OP

Thank you.

Answer 1

Set $t=\ln x$ and $dt=\frac{dx}{x}$

$$=\int\frac{dt}{\sqrt{t^2+6t}}\overset{\text{complete the square}}{=}\int \frac{dt}{\sqrt{(t+3)^2-9}}$$

Set $s=t+3$ and $ds=dt$

$$=\int\frac{ds}{\sqrt{s^2-9}}$$

Set $s=3\sec(p)$ and $ds=3\tan p \sec p dp$, then $\sqrt{s^2-9}=\sqrt{9\sec^2 p-9}=3\tan p$ and $p=\sec^{-1}\left(\frac s 3 \right)$

$$=\int \sec pdp=\ln(\tan p +\sec p)+\mathcal C=\dots$$

Answer 2

Once you have the differential in the form $\frac{ds}{\sqrt{s^{2}-a^{2}}}$, the integral will be just the inverse hyperbolic cosine of $\frac sa$ plus a constant.