Integration?

Question

How do you integrate:
t^2 / (1+t^3)
Thanks a lot.

Answer 1

u = 1+t^3
du = 3t^2 dt

∫[t^2 / (1+t^3)]dt
= ∫(1/3)(1/u)du
= (1/3)ln|u| + c
= (1/3)ln(|1+t^3|) + c

Answer 2

substitute (1+t^3) = u
then 3(t^2) * dt = du,
or (t^2) * dt = 1/3 * du

so integral {t^2 / (1+t^3)} dt
= integral {1/3 * 1/u} du
= 1/3 * ln (u) + constant
= 1/3 * ln (1+t^3) + constant ...Final answer

Answer 3

∫t^2 / (1+t^3) dt
= ∫(1/3) / (1+t^3) d(1+t^3), mental substitution
= (1/3)ln|1+t^3| + c

Answer 4

Use "u substitution"
set u equal to denom ---- u = 1 + t^3
take deriv w/ respect to t ---- du = 3t^2 dt
get dt by it self ---- du/3t^2 = dt

Original problem ----int (t^2/(1 + t^3) * dt)
now "substitute" u and dt in---- int (t^2/(u )* du/3t^2)
t^2 cancel out ----- int (1/(u )* (1/3)du)
pull out 1/3, look familiar? ----- (1/3) int (1/(u )* du) =(1/3) Ln U +c
substitute u back in and get ---- (1/3)Ln( 1 + t^3) + C ---- thats your answer!

Answer 5

I = ∫ t² / (1 + t³) dt
I = (1/3) ∫ 3 t ² / (1 + t ³) dt
I = (1/3) log (1 + t ³) + C

Answer 6

[ ln (1+t^3) ] / 3 + c

Answer 7

answer: [ ln (1+t^3) ] / 3 + c