Integration by substitution

1. I = ∫[(2x+3)/(x+2)] dx
Let x + 2 = udx = du2x + 3 = 2(x + 2) – 1 =2u – 1I = ∫[(2u – 1)/ u] du= ∫[2 – 1/u] du= 2u – ln |u| + C= 2x – ln |x + 2| + C’ where C’ = C + 42. I = ∫[1/xln(x^2)] dxLet u = ln(x^2)du = 2x/x^2 dx = 2dx/x => dx/x = du/2Hence I = 1/2∫[1/u] du=1/2 ln u + C= 1/2 ln[ln(x^2)] + C

2011-03-13 22:56:07 補充：
1/2 ln | ln|x^2| | + C
= 1/2 ln| 2ln|x| | + C
=1/2 ln2 + 1/2 ln| ln|x| | + C
= 1/2 ln| ln|x| | + C'

It's better to let u to solve the integrals.

2011-03-13 22:16:49 補充：
For Q2, the model answer is quite different from yours.
the answer is 1/2 ln│ln│x││+ C, but I don't know why is like that.