integration of t^2/9+t^6 dt
ok...now i have (du/3)/9+u^2 and i have no idea where to go next...i could pull out a (1/3) in the front but then what? i also know that inverse tan is 1+u^2 which is something i'm looking out for to make the equation work...thanks for all the help by the way...
you have $\displaystyle \frac 13 \int \frac {du}{9 + u^2}$ so far
now we want to get the denominator to the form $\displaystyle 1 + x^2$ so we can use the arctan for the integral. the 9 is the problem, so factor it out and simplify
$\displaystyle \frac 13 \int \frac {du}{9 + u^2} = \frac 13 \int \frac {du}{9(1 + u^2/9)}$
$\displaystyle = \frac 1{27} \int \frac {du}{1 + (u/3)^2}$
now you may want to do a second substitution here of $\displaystyle x = \frac u2$, to get: $\displaystyle \frac 2{27} \int \frac {dx}{1 + x^2}$