These few questions have been bugging me for hours, help would be appreciated
Because with that subsitution you can 'eliminate' the root form which is easier to work with. (Have you solved the integral?)
$\displaystyle \int \frac{9t}{9t^2+6t+5}dt = \frac{1}{2} \int \frac{(18t+6)-6}{9t^2+6t+5}dt = \frac{1}{2} \left(\int \frac{18t+6}{9t^2+6t+5}dt - \int \frac{6}{9t^2+6t+5}dt\right)$
I guess you can take it from here. To solve $\displaystyle \int \frac{6}{9t^2+6t+5}dt$ write $\displaystyle 9t^2+6t+5=(3t+1)^2+4$ (because $\displaystyle D<0$)
$\displaystyle sin^2u+ cos^2u= 1$ so, dividing both sides by $\displaystyle cos^2 u$, $\displaystyle tan^2u+ 1= sec^2u$. Because of that the substitution $\displaystyle x= tan u$ changes $\displaystyle \sqrt{1+ x^2}$ to $\displaystyle \sqrt{1+ tan^2 u}= \sqrt{sec^2 u}= sec u$. Generally speaking, any time you have a square root of a quadratic, you should consider a trig substitution. (Some people prefer hyperbolic functions which satisfy similar identities.)
As for your new question, first complete the square: $\displaystyle 9t^2+ 6t+ 5= 9(t^2+ (2/3)t+ 1/9- 1/9)+ 5)= 9(t+ 1/3)^2+ 4$.
Now the substitution x= t+ 1/3 changes to integrand to $\displaystyle \frac{9u- 3}{9u^2+ 4}= \frac{u}{u^2+ (2/3)^2}+ (4/9)\frac{1}{(3u/2)^2+ 1}$