Integrate 1 / x(2sqrt(x)-1) on [1,9] using x = u^2 (u > 0).

Differentiate x = u² to get dx = 2u du. We need to change the limits, too:

1 <= x <= 9  <==>  1 <= u² <= 9  <==>  1 <= u <= 3  (since we are given u > 0).

Now we can substitute in the integrand:

dx / x (2sqrt(x) - 1) = (2u du) / u²(2u - 1) = (2 du) / u(2u -1).

Noticd that we can write 2 / u(2u - 1) = 4 / (2u -1) - 2 / u, so that

Integral(2 / u(2u-1)) du = Integral( 4 / (2u - 1) ) du - Integral( 2 / u ) du

                                      =  2 ln(2u-1) - 2 ln(u) + c.

The value of the definite integral is 2 ln (5/6), which follows by a simple calculation in the above.