Evaluate the integral, if it exists:?

Question

The integral (with lower limit 0 and upper limit 1) of (4sqrt of u + 1)^2 du.
Note: The 4 is part of the square root, not 4 times the sqrt of u.

Yes, I meant the fourth root of u. Thanks!

Answer 1

∫u = 0 to 1 (u^(1/4) + 1)^2 du
= ∫u = 0 to 1 (u^(1/2) + 2u^(1/4) + 1)du
= [2/3 u^(3/2) + 2 * 4/5 u^(5/4) + u] u = 0 to 1
= [2/3 + 8/5 + 1] - [0]
= 49/15

Answer 2

When you say the 4 is part of the square root,
do you mean the 4th root of u? The question
is not clear. Sorry!