Quick question about trigonometric substitution $\int\frac{u+5}{u^2+9}du$

Question

I'm trying to calculate this integral::

$$\int\frac{u+5}{u^2+9}du$$

which is a part of:

$$\int\frac{e^x}{(e^x-5)(e^{2x}+9)}dx$$

The part I posted is the only one giving me a wrong answer. I am using trigonometric substitution, but I can't see the error I've made so I'd like someone to point out what's wrong with my method:

Using $u=3\tan\theta$

$$\int\frac{u+5}{u^2+9}du$$ $$\int\frac{(3\tan\theta+5)(3\sec^2\theta)}{9\tan^2+9}d\theta$$ $$\int\frac{(3\tan\theta+5)(3\sec^2\theta)}{9\sec^2\theta}d\theta$$ $$\frac{1}{3}\int(3\tan\theta+5)d\theta$$ $$\int(\tan\theta)d\theta +\frac{5}{3}\int(d\theta)$$ $$\log|\sec\theta| + \frac{5}{3}\arctan\frac{u}{3}$$ $$\log|\frac{\sqrt{u^2+9}}{3}| + \frac{5}{3}\arctan\frac{u}{3}$$

Apparently, the logarithm is incorrect.

Thank you for the help.

Answer 1

Hint: Split this into two integrals.
$$\int\frac u{u^2+9}du+5\int\frac{du}{u^2+9}$$

The first doesn't require trig substitution, just $u$-substitution. Use trig substitution on the second one.

Answer 2

$$\int\frac{u+5}{u^{2}+9}du\\ u=3v\Rightarrow du=3dv\\ \int\frac{u+5}{u^{2}+9}du=\int\frac{3v+5}{9v^{2}+9}3dv=\frac{1}{2}\int\frac{2vdv}{v^{2}+1}+\frac{15}{9}\int\frac{dv}{v^{2}+1}\\ \int\frac{u+5}{u^{2}+9}du=\frac{1}{2}\ln(\frac{u^{2}}{9}+1)+\frac{5}{3}\arctan(\frac{u}{9})+c$$