Question

integral from 1 to t^2 of (1+u^4)^.5/u

Answer 1

why am I seeing 3 variables in your integral and no differential o.O ?

Answer 2

what is u ?
is this integral with respect to x or what?

Answer 3

@TuringTest sorry about that the (1+x^4) is actually (1+u^4)

Answer 4

\[\int\limits_{1}^{t^2}\frac{ \sqrt{1+u^4} }{ u }\]

Answer 5

I'm thinking trig sub, but it seems strange that the upper bound is t^2
the problem is not\[\frac d{dt}\int_1^{t^2}{\sqrt{1+u^4}\over u}du\]is it?

Answer 6

The red number doesn't matter, but here is the problem

Answer 7

neither red number matters? the 2 and the 4 are irrelevant you say?

Answer 8

use basic properties of antiderivatives

Answer 9

if you decide to find f'(x) then we are finding dx/dt, which will remove the integral on right side of equation

Answer 10

infact there is no need to use any formulae

Answer 11

I think I understand the problem now, and you do at least need FTOC

Answer 12

by the fundamental theorem of calculus\[F(x)=\int_1^x f(t)dx\implies F'(x)=f(x)=\int_1^{x^2}\frac{\sqrt{1+u^4}}udu\]this means that\[F''(x)=f'(x)=\frac{\sqrt{1+x^8}}{x^2}\cdot(2x)=\frac{2\sqrt{1+x^8}}x\]does this make sense to you?

Answer 13

thank you so much, that really does, I had found the derivative of that, so I kept on getting it wrong.

Answer 14

yeah you have to use the chain rule on the t^2 thing, and the double application of FTOC is a bit confusing
interesting prob though :)