Integrals

[Frogger888]

I have a question. I am having trouble solving for integrals that involve ln as a final solution for example

xdx/[(x-1)^2] is ln[x-1] - 1/(x-1)

what are the rules involving ln as a solution??? I am in Diff Eq but the calculus is what is difficult.

How do you go about getting answers for such solns.

Another example. The integral of 3/(x-2)

Please help explain this
Thanks

 

[daveyp225]

x/(x-1)^2 dx

u = x-1
du = dx
x = u+1

substitute (u+1) / u^2 du

expand by numerator ( u/u^2 + 1/u^2 ) du

simplify (u^-1 + u^-2) du

distribute u^-1 du + u^-2 du

Now, the rule for Ln says that if you have (u)^-1 * du then the integral is Ln|u|. The second part is strictly the power rule

Ln|u| + (u^-1) / -1 => Ln|u| - u^-1

Substitute back in for u: Ln|x-1| - (x-1)^-1.

*** My suggestion to you is to buy a good calc textbook if you don't have one already and study the inside/back covers to get all the rules down pat. If you get confused with one of the rules, a good book will have the explanation and proof in it.

For the second problem, set u = x-2. Then du=dx