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Struggling with the transcendentals

  1. Feb 8, 2013 #1
    I'm three weeks in to my Calculus II semester and we've just finished covering derivatives and integrals of [itex]ln[/itex] and [itex]e^x[/itex]. We haven't had the first test yet, which will be covering this material.

    Things that trip me up are some of the tricky u-substitution methods, change of form, etc., or one in particular is: [itex]\int\frac{2x}{(x-1)^2}dx[/itex]

    A couple of classmates were working together the other day and the way one was showed to work this problem out was by changing the numerator to [itex]2(x-1)+2[/itex] which is essentially [itex]2x[/itex]. The thing that gets me about this is, I have no idea how to look at that example and know that I need to perform that change of form, my mind doesn't even see that.

    Mostly I feel like I'm getting the calculus concepts, aside from what I've mentioned above, however I have always been a bit shaky with algebraic skills. It seems most of my mistakes come from making "simple" errors or not seeing ways of simplifying something that would be obvious to someone with a more solid understanding of algebra.

    It's an odd feeling. If I know what problems I'm having, it should be easy to go back to certain topics in algebra and build up on them, but it hasn't worked out that way when I try it.

    I thought someone might have the right tip for me, maybe trying something different will help me progress a little easer. Until then, I'll keep working examples, as that's the only thing I know to do to get better at these.
  2. jcsd
  3. Feb 8, 2013 #2
    You choose u based on the denominator and you need all the x's to go away in the integrand. Since u = x+1, you need to express everything in terms of (x+1) so that cancellation occurs and all x's are out of the integrand.
  4. Feb 8, 2013 #3
    Do you mean x-1?

    Anyway, in calculus, adding and subtracting or multiplying and dividing the same quantity are commonplace. Try to think of what you can add/subtract/multiply/divide to simplify the expression, and then make a substitution on what's left.
  5. Feb 8, 2013 #4
    Thanks, I definitely see how in the example I used, doing what you say does make it simpler to work with. I'll try to get used to thinking this way more often. Maybe this is whats giving me difficulty, I'm not used to thinking this way to solve a problem mathematically..
  6. Feb 8, 2013 #5


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    Perhaps a little more direct way to look at it is to note that [itex]\int 2x/(x-1)^2 dx[/itex] and, generally, the more complicated a denominator is, the harder the integral will be. So Let u= x- 1, du= dx. Now, from that, x= u+ 1 and so 2x= 2u+ 2.

    The integral becomes [itex]\int (2u+ 2)/u du= 2\int du+ 2\int (1/u)du[/itex].
  7. Feb 8, 2013 #6
    One more way to solve it is integration by parts.

    [itex]\int u(x)v'(x)dx[/itex] = [itex] u(x)v(x) - \int u'(x)v(x)dx[/itex]

    Substituting [itex]u(x) = 2x[/itex] and [itex]v(x) = - \frac{1}{(x-1)} [/itex]:

    [itex] \int \frac{2x}{(x-1)^2}dx = - \frac{2x}{x-1} + \int \frac{2}{x-1}dx = - \frac{2x}{x-1} + 2ln(x-1).[/itex]
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