Don't know how to answer

Question

Given that ∫1 to 0 9x√x^2+9 dx=30√10−81, what is ∫0 to 1 9u√u^2+9 du?

Answer 1

∫_{(0 to 1)} [9x √(x²+9)] dx

Let u = x² + 9. Then du = 2xdx. So, xdx = (1/2)du.

When x = 0, u = 9 and when x = 1, u = 10.

So, we have 9∫_{(9 to 10)} √u (1/2)du = (9/2)(2/3)u^3/2_{(from 9 to 10)}

= 3u^3/2_{(from 9 to 10)} = 3[10^3/2 - 9^3/2] = 3(10√10 - 27) = 30√10 - 81.