using substitution wrongly Solving integral, first way:
$$\int \frac{du}{u^2-9}=-\int \frac{du}{9-u^2}$$
$$u={3\sin v}$$
$$du=3\cos vdv$$
$$-\int \frac{3\cos vdv }{9-9\sin^{2}v}=-\frac 13\int\frac{dv}{\cos v}=-\frac 13\ln\left(\sec v+\tan v\right)=-\frac13 \ln \frac {\frac u3}{\sqrt{1-\frac{u^2}{9}}}$$
$$-\frac13 \ln \frac{u}{\sqrt{9-u^2}}=-\frac16 \ln \frac {u^2}{9-u^2}=\frac16 \ln \frac{9-u^2}{u^2}$$
second way:
$$\int \frac{du}{u^2-9}=\frac 16\int \frac{du}{u-3}-\frac16 \int \frac{du}{u+3}=\frac16 \ln \frac {u-3}{u+3}$$
what's wrong with first way?
Thanks all. $1$ from $\sec$ is missed, must be $1+\frac u3$

As Grid commented, before any simplification
$$\ln\left(\sec v+\tan v\right)=\log \left(\frac{u}{3 \sqrt{1-\frac{u^2}{9}}}+\frac{1}{\sqrt{1-\frac{u^2}{9}}}\right)$$
