Question

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

Answer 1

Your equation is \[t^2(\frac{t}{t-8})\]?

Answer 2

\[t^2(t-\frac{8}{t})\]

Answer 3

?

Answer 4

\[\int\limits_{?}^{?} t^{2}(t-\frac{ 8 }{ t })\]

Answer 5

But this is just t^3 - 8t

Answer 6

nooo.. i have the answer from the book and its \[\frac{ 1 }{ 4 }x^{4}-4x^{2}+c\]

Answer 7

Which is exatcly the antiderivative of t^3 - 8t

Answer 8

i have to use u substitution....
you know what that is

Answer 9

Yes. By why use it if you can calculate it just as easily with out it?

Answer 10

because teacher wants it that way

Answer 11

\[\int t^2\left(t-\frac{8}{t}\right)~dt\]
Letting \(u=t-\dfrac{8}{t}\) gives \(du=\left(1+\dfrac{8}{t^2}\right)~dt\).

Hmm... notice that \(u=t-\dfrac{8}{t}=t\left(1-\dfrac{8}{t^2}\right)\), but that doesn't really help... What you need is an expression for \(t\) or \(t^2\) in terms of \(u\), but it doesn't look like that's possible with the given substitution.

Sorry, but I don't see a way a substitution can help here, unless you mean something else by that integral.

Answer 12

you could try
u = t^2
du = 2t dt

\[\int\limits t^2 (t - \frac{8}{t}) dt = \int\limits t^2 (\frac{t^2 -8}{t})dt = \frac{1}{2} \int\limits (u -8) du\]