Integral of x/(x^4+1) with respect to t

[brandy]

Homework Statement

integral of x/(x^4+1) with respect to t

Homework Equations

i know your spose to use (x^2)^2=x^4, and use chain rule
but apart from that, i have ABSOLUTELY no idea.

The Attempt at a Solution

i have no idea where to begin, i tried doing it my reverse product rule but that was an epic fail. if i did i woudlnt be asking for help here.

 

[HallsofIvy]

Yes, write this as $x/((x^2)^2+ 1)$ and use the substitution $u= x^2$. Then du= 2x dx of (1/2)du= xdx. The denominator becomes u^2+ 1 so you are now integrating (1/2) du/(u^2+ 1) which should be an easy integral.

 

[tiny-tim]

Hi brandy!

(try using the X² tag just above the Reply box )

integral of x/(x^4+1) with respect to t

If u = x², then du = … ?

 

[brandy]

keep in mind I am really REALLY slow.
is this right, you said du= 2x *dx
so du/2=x*dx
and u subbed that in
for the numerator?
if so, where did the dx go

 

[tiny-tim]

… where did the dx go

∫ x dx / (x⁴ + 1)

= ∫ (1/2) du / (x⁴ + 1)

= … ?

 

[brandy]

ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo i get it now

 

[tiny-tim]

Wooooooohooooooooooooooooo! ​

 

[Feldoh]

I might be missing something but doesnt

integral of x/(x^4+1) with respect to t = $$\int \frac{x}{x^4+1} dt$$

 

[Mark44]

I might be missing something but doesnt

integral of x/(x^4+1) with respect to t = $$\int \frac{x}{x^4+1} dt$$

Or maybe Brandy really means something like this:
$$\int_0^t \frac{x dx}{x^4 + 1}$$

If that's the case, then this would better be described, in words, as the definite integral from 0 to t of x divided by x⁴ + 1.

If that's not the case, then I have no idea what the problem is.

 

[brandy]

lol my bad. i was meant to say with respect to x not t