1. ## integral

$

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

$

it says to take the the u substitution of $\frac{1}{3}t^3$ 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.

$u = \frac{1}{3}t^3
$

$du = t^2 dt$

so the integral becomes

$\int \frac{t^2}{3^2+{3u}^2} \frac{du}{t^2}
$

$\int \frac{1}{3^2+{3u}^2} du
$

$= \frac{1}{9} \arctan u
$

$= \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 $u = \frac{1}{3}t^3 \ \ \Rightarrow \ \ 3u = t^3$

also as $du = t^2 dt \ \ \Rightarrow \ \ dt = \frac{du}{t^2}$

so we have

$\int \frac{t^2}{3^2+(t^3)^2} dt$ we replace $dt$ with $\frac{du}{t^2}$ and $(t^3)^2$ with $(3u)^2$

giving
$\int \frac{t^2}{3^2+{3u}^2} \frac{du}{t^2}
$

do you follow the rest ?