what is the integral of ((t^4)+1)^(1/2)???

Question

please tell me how you did it and the name of the procedure or theorem

Answer 1

You have to think backwards to get this. It takes practice. Try to think, "What turns into this function?"

Ex: If I was integrating "D^2" I would use my calculus knowledge to know to multiply "D^2" by (1/3) and then raise the power of (D^2)one (3) to get the answer.

When you derive (1/3) X (D^3), the (1/3) out in front is multiplied by the power of "#" in "D^#" and then the power on D is lowered by subtracting one from it to get (D^2). The 3 X (1/3) = 1. We're back where we started so we know that (D^2) was integrated right. I hope this helps.

Here is the math answer to your question:

(2/3) X ((1/4)(t^-3)) X ((t^4)+1)^(3/2) =

(1/6) X (t^-3) X ((t^4)+1)^(3/2) (simplified form)

X= multiply

(t^-3) = (1/(t^3))

Answer 2

Sorry, this integral cannot be evaluated in terms
of elementary functions. Since you are
integrating the square root of a quartic polynomial,
you have an elliptic integral.
Wolfram Math Integrator reduces this
to an elliptic integral of the first kind.
Stick-jockeyman's answer is incorrect.