$\displaystyle
\int \frac{t^2{dt}}{9+{t^6}}
$
it says to take the the u substitution of $\displaystyle \frac{1}{3}t^3$ but how do u get that and is there a simpler way of doing this
the given substitution is a very simple way of doing it.
$\displaystyle u = \frac{1}{3}t^3
$
$\displaystyle du = t^2 dt$
so the integral becomes
$\displaystyle \int \frac{t^2}{3^2+{3u}^2} \frac{du}{t^2}
$
$\displaystyle \int \frac{1}{3^2+{3u}^2} du
$
$\displaystyle = \frac{1}{9} \arctan u
$
$\displaystyle = \frac{1}{9} \arctan \frac{1}{3}t^3
$
I omitted a few step of working. let me know if you can't follow that.
Bobak
as $\displaystyle u = \frac{1}{3}t^3 \ \ \Rightarrow \ \ 3u = t^3$
also as $\displaystyle du = t^2 dt \ \ \Rightarrow \ \ dt = \frac{du}{t^2}$
so we have
$\displaystyle \int \frac{t^2}{3^2+(t^3)^2} dt $ we replace $\displaystyle dt$ with $\displaystyle \frac{du}{t^2}$ and $\displaystyle (t^3)^2$ with $\displaystyle (3u)^2$
giving
$\displaystyle \int \frac{t^2}{3^2+{3u}^2} \frac{du}{t^2}
$
do you follow the rest ?