Well, I've looked these over, and I have a solution to number 1, but sorry, I don't know how to do number 2.
1. /int 2_1 (e^(1/x)/x^2)dx the answer is e-e^(1/2)
2. /int pi/6_0 tan 2x dx the answer is (1/2)ln 2
3. I have a question, for a definite that have the same number except one is positive and the other is negative (/int 2_-2), is there a special way or equation to find the answer? It goes something like /int 0_-2 (-anything)-/int 2_0 (anything) or something of the like. I hope this is enough infomation.
For 1.u=1/x, du=-1/x^2. Well, in this circumstance, I don't know the rest since if it needs to be multiplied or divided or not.
For 2. u=2x du=2 dx. tan u -> -ln|cos(u)|*(1/2) and i think this is where i may have gotten something wrong.
So if we have an integral from a to b of some function of x (say f(x)) with repect to x and we make the change of variable u=h(x), the new integral is with respect to u from u=h(a) to u=h(b).