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Finding the antiderivative

  1. Mar 30, 2006 #1
    My AP calculus course has just started integration and so on. Because the exam date does not match up with my school's schedule, I'm kind of learning all of this in a disjointed manner at a faster rate than a normal class...

    I'm starring at this question, and am drawing a blank:

    find the antiderivative of f
    [tex]f (x)=4-3{(1+x^2)}^{-2}[/tex]
    Last edited: Mar 30, 2006
  2. jcsd
  3. Mar 30, 2006 #2


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    Use the linearity of the integral, the 4 shouldn't be a problem then.
    For the second part, you can consider working with the rational fraction (integration by parts or whatever) of using the substitution x = tan(y) to use 1+tan²y = sec²y.
  4. Mar 30, 2006 #3


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    Do you now what function has derivative 4? do you know what function has derivative x-2? Can you apply the chain rule to this?
  5. Mar 30, 2006 #4
    TD: I think I know where you are going with this, but the thing is, I haven't covered the derivatives of trigonometric functions in depth yet.. like I said, I'm kind of going through the course in a disjointed manner. We're kind of covering the textbook simultaneously from both ends and meeting somewhere in the middle, where all the trigonometry is. I've looked at the section, but not in enough detail to be finding their integrals and so on. :frown:

    HallsofIvey: I'm not quite sure of what you mean by . Could you please elaborate?

    I suppose it'd help you guys to help me if I try to explain what I'm trying to do and why I'm getting stuck.

    Taking the antiderivative of 4 is no big deal; 4x. For the second part of the fuction, I orignally tried to apply the chain rule in reverse, and went:

    [tex]4x+(\frac{3}{2x}) {(1+x^2)}^{-1} [/tex]

    However, I knew that this answer was wrong once I looked at it because if I tried to differentiate this function, I would have to use the product rule for

    [tex](\frac{3}{2x}) {(1+x^2)}^{-1} [/tex]

    Even though I know what I did was incorrect, I really can't see another way of solving the problem on my own. =\
    Last edited: Mar 30, 2006
  6. Mar 30, 2006 #5
    I think you'll either have to do it by parts, or through substitution as TD said. I can't see any other easy way of doing it.
  7. Mar 30, 2006 #6


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    If you're uncomfortable with the trig substitution, there is indeed still the option of integration by parts. Have you covered that yet?
  8. Mar 30, 2006 #7
    HallsofIvy simply made a typo - he forgot a forward slash. He meant x-2.

    Say you want to say x^2. You can actually put the exponent in like this: x2. You have to use [ sup ] and [ / sup] (no spaces of course). So you would type x[ sup ]2[ / sup]

    To say x0, you use [ sub ] and [ / sub], so you would type x[ sub ]0[ / sub]. It's very handy =)
  9. Mar 30, 2006 #8
    We've just covered integration by parts last lesson, but I'm not entirely sure how I could apply it to my given situation.

    In regards to HallsofIvy's suggestion to finding the antiderivative of [tex]x^2[/tex] and applying the chain rule, I've already tried that and ran into problems, as I stated earlier =\

    Thanks a lot for the help; this must be as frustrating for you guys as trying to teach a deaf child sing.
  10. Mar 30, 2006 #9


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    Could you clarify the problem? In your first post there's a 1+x² and further on it reads 1-x² in the denominator.
  11. Mar 30, 2006 #10
    oops, that's supposed to be [tex]1+x^2[/tex] in both cases.

    I've edited my previous post with the typo.
  12. Mar 30, 2006 #11


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    Ok, then consider this trick:

    \int {\frac{1}
    {{\left( {1 + x^2 } \right)^2 }}dx} = \int {\frac{{1 + x^2 - x^2 }}
    {{\left( {1 + x^2 } \right)^2 }}dx} = \int {\frac{{1 + x^2 }}
    {{\left( {1 + x^2 } \right)^2 }}dx} - \int {\frac{{x^2 }}
    {{\left( {1 + x^2 } \right)^2 }}dx} = \int {\frac{1}
    {{1 + x^2 }}dx} - \int {\frac{{x^2 }}
    {{\left( {1 + x^2 } \right)^2 }}dx}

    Now the first integral is a basic integral (arctan(x)). Can you use integration by parts on the second?
  13. Mar 30, 2006 #12
    Can we solve this
    \int {\frac{{x^2 }}{{\left( {1 + x^2 } \right)^2 }}dx} [/tex]

    using partial fractions
    Last edited: Mar 30, 2006
  14. Mar 30, 2006 #13
    I don't think so, as [tex] \frac{x^2}{(1+x^2)^2} = \frac{1}{1+x^2} - \frac{1}{(1+x^2)^2} [/tex]

    Won't we be back to where we started?
  15. Mar 31, 2006 #14


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    Indeed, partial fraction won't work but integration by parts will.
  16. Mar 31, 2006 #15


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    Or one can try to do the following if you don't want a trig substitution:
    [tex]\int \frac{dx}{1 + x ^ 2}[/tex].
    Integrating by Parts gives:
    [tex]\int \frac{dx}{1 + x ^ 2} = \frac{x}{1 + x ^ 2} + \int \frac{2x ^ 2}{(1 + x ^ 2) ^ 2} dx = \frac{x}{1 + x ^ 2} + \int \frac{2x ^ 2 + 2 - 2}{(1 + x ^ 2) ^ 2} dx = \frac{x}{1 + x ^ 2} + \int \frac{2}{(1 + x ^ 2)}dx - \int \frac{2 dx}{(1 + x ^ 2) ^ 2}[/tex].
    [tex]\Leftrightarrow - \int \frac{dx}{1 + x ^ 2} = \frac{x}{1 + x ^ 2} - 2 \int \frac{dx}{(1 + x ^ 2) ^ 2}[/tex].
    So what's:
    [tex]\int \frac{dx}{(1 + x ^ 2) ^ 2} = ?[/tex].
    Can you go from here? :)
    Last edited: Mar 31, 2006
  17. Mar 31, 2006 #16
    Hey guys, thanks a lot for the suggestions. I ended up solving the thing by integrating in parts. Thanks for all the other hints too, it really showed me some ways of solving the problem that I didn't even think of.
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