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.
TD
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