find the integral with using u substitution, integral t^2 (t-8/t) u=(t-(8/t)) and du=t^2. now i understand that you find u integral, but i am having trouble with tying it with du please help

Question

kainui · Answer

So this is your integral? $$\Large \int\limits t^2(t-\frac{8}{t})dt$$

anonymous · Answer

yes

kainui · Answer

And you have to use u=(t-8/t)? It seems much better to just distribute the t^2 to the terms to make the integral into $$\Large \int\limits t^3-8t \ dt$$

anonymous · Answer

ok i didnt see that, never thought of distributing

kainui · Answer

Cool.

anonymous · Answer

so then you would have $$\frac{ 1 }{ 4}\int\limits_{}^{} \frac{ u ^{2} }{ 2}$$
correct right?

kainui · Answer

No, you don't have to do any substitution at all. You simply just do the opposite of differentiation here.

anonymous · Answer

but i have to use it

kainui · Answer

Why?

anonymous · Answer

i have to to show my work

kainui · Answer

Well do you know this? $$\Large \int\limits x^ndx=\frac{x^{n+1}}{n+1}$$

This is just the reverse of differentiation. This is all that's left to do. A substitution wouldn't make any sense to me here.

anonymous · Answer

yes. if i am correct it would be $$\frac{ t^{4} }{ 4 }$$

kainui · Answer

However that's not the entire thing, you still have -8t to integrate as well! Don't forget the constant of integration like I did! =P

anonymous · Answer

yeah, i did i just didnt type it

anonymous · Answer

ok i see, i was just seeing it in one way

anonymous · Answer

thank you