1. ## integral

$\displaystyle \int \frac{t^2{dt}}{9+{t^6}}$

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

2. 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

3. how did you get the very first step of the integral....

4. 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 ?