# Thread: Find the anti-derivative

1. ## Find the anti-derivative

integration of t^2/9+t^6 dt

2. Originally Posted by simsima_1
integration of t^2/9+t^6 dt
this is a substitution problem (you may want to do a substitution twice, but you don't have to). let u = t^3

3. if i use t^3 as my u will that affect the (9+t^6)?

4. Originally Posted by simsima_1
if i use t^3 as my u will that affect the (9+t^6)?
yes. 9 + t^6 = 9 + (t^3)^2 = 9 + u^2

5. 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...

6. Originally Posted by simsima_1
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 $\frac 13 \int \frac {du}{9 + u^2}$ so far

now we want to get the denominator to the form $1 + x^2$ so we can use the arctan for the integral. the 9 is the problem, so factor it out and simplify

$\frac 13 \int \frac {du}{9 + u^2} = \frac 13 \int \frac {du}{9(1 + u^2/9)}$

$= \frac 1{27} \int \frac {du}{1 + (u/3)^2}$

now you may want to do a second substitution here of $x = \frac u2$, to get: $\frac 2{27} \int \frac {dx}{1 + x^2}$

7. i made a typo in my last post. i typed let x = u/2 when it should be x = u/3. make the necessary corrections