What is the antiderivative of dt/(1+t^4) ?

Question

I thought I had to use a natural log or something, but I'm confused.

thanks.

it's for an integral problem

its

f(x) = (integral symbol: x on top, 1 on bottom) dt/(1+t^4).

Then asks, what is f'(2)?

Answer 1

It seems you are asking for f'(2)? in which case, we have
f'(x) = 1/(1+x^4)so that f'(2) = 1/9

Steve

Answer 2

The main manipulation is in the first step!
∫dt/(1+t^4) = (1/2)∫[(2+t^2-t^2)/(1+t^4)]dt

=>(1/2)∫[(2+t^2-t^2)/(1+t^4)]dt= (1/2)∫(1+t^2)dt/(1+t^4)
+ (1/2)∫(1-t^2)dt/(1+t^4)

Let the first integral be A and 2nd one be B, dividing numerator and denominator in both the integrals by t^2

=>A=(1/2)∫(1/t^2 + 1)dt/(t^2 + 1/t^2)
B=(1/2)∫(1- 1/t^2)dt/(t^2 + 1/t^2)

solving A
A=(1/2)∫(1/t^2 + 1)dt/(t^2 + 1/t^2)

=>A=(1/2)∫(1/t^2 + 1)dt/(t^2 + 1/t^2+1-1)

=>A=(1/2)∫(1/t^2 + 1)dt/((t^2 + 1/t^2-1) + 1)

=>A=(1/2)∫(1/t^2 + 1)dt/((t -1/t)^2 + 1)

let t - 1/t= p
=>(1+1/t^2)dt= dp

=>A=(1/2)∫dp/(p^2 + 1) = (1/2)Arctan(p) + C
Substitute p=t-1/t

=>A=(1/2)Arctan(t-1/t) + C

Now solve for B

B=(1/2)∫(1- 1/t^2)dt/(t^2 + 1/t^2)

=>B=(1/2)∫(1- 1/t^2)dt/(t^2 + 1/t^2 +1 -1)

=>B=(1/2)∫(1- 1/t^2)dt/[(t + 1/t)^2 -1]

let t+1/t= y
=>(1-1/t^2)dt= dy

=>B=(1/2)∫dy/[y^2 -1] = (1/2)* (1/2)log|(y-1)/(y+1)| + c

substitute y=t+1/t
=>B=(1/4)log|(t+1/t + -1)/(t+1/t - 1)| + c


=>A+B=(1/2)Arctan(t-1/t) + (1/4)log|(t+ 1/t + -1)/(t+1/t - 1)| + K


results used
1)∫dx/(x^2+ a^2)= (1/a)Arctan(x/a) + C
2)∫dx/(x^2 - a^2)=(1/(2a))log|(x-a)/(x+a)| + C

Answer 3

I'm not a math guy, but I think you have to solve the integral.
Once solved you will have tu substitute the x by the value 2. And then solve the rest of the ecuation.

That is the integral of a curve between two points, the poits are X and 1, but since they are asking you to solve it for f'(2) I guess you will have to solve the integral between 2 and 1.