Right?
The point behind the substitution is to rewrite the integrand in terms of u only. So
and
which gives
Now you have a value for both t and dt to put into your integrand.
This substitution doesn't help you do the problem, I just wanted to run you through how to set it up properly.
-Dan
This is the method I outlined a couple of days ago.
Let the integral equal and let (to avoid various fractional powers of 5 complicating the calculation).
Then
Split the integrand into two (partial) fractions.
Let
then
Substituting and equating coefficients of and constants across the identity leads to
Substituting in and tidying up a little produces
Rewrite the top line of the second fraction as and split this fraction into two, then
Complete the square in the denominator of the third fraction and introduce the factor top and bottom to arrive at
Finally, integrate to get
a constant of integration (and where is the cube root of 5).