Evaluate the integral from 2/sqrt2 to 4; 1/(t^3 sqrt(t^2-4))?

Answer 1

The indefinite integral is
[1/(8t^2)]*sqrt(t^2-4) + (1/16)cos^-1(2/t)

I got it from a table of integrals.

Evaluate it at t = 4 and t = sqrt2
and take the difference.

Note that 2/sqrt2 = sqrt2.

There is a problem with your limits of integration. The function is undefined for t < 2, and yet you are trying to start the integration at sqrt2