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Indefinite Integrals

  1. Aug 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Evaluate the Integrals:

    [tex]\int \frac{-2}{\sqrt{1 - x^2}} dx[/tex]

    [tex]\int \frac{2x+5}{x^2+6x-3} dx[/tex]

    [tex]\int \frac{x^3}{x^2-1} dx[/tex]

    [tex]\int \frac{x}{1+x^4} dx[/tex]

    Verify that this integral is correct:
    [tex]\int ue^a^u du = \frac{e^a^u}{a}(u-\frac{1}{a})+C[/tex]
    Last edited: Aug 9, 2007
  2. jcsd
  3. Aug 8, 2007 #2


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    Staff: Mentor

    Welcome to the PF, alternet87. One of the rules here is that you must show some of your own work before we can offer tutorial advice. We do not do the work for you.

    So how would you start your first integral?

    [tex]\int \frac{-2}{\sqrt{1 - x^2}} dx[/tex]
  4. Aug 8, 2007 #3
    I have no idea, that is why I need your help.

    maybe, drop the -2 down because the integral of 1/sqrt(1-x^2) is cos(x) or sin(x). but what happens to the -2 ??? please help

    for the second one maybe substituting u in for the x^2 +6x-3. Then du=2x+6 so x=(1/2)du-3. and putting that in for x would give us:
    S (1/u)*2*((1/2)du-3)+5
    S -1+(1/u) du
    = ln(u) - u
    = ln(x^2 +6x-3) - (x^2 +6x-3)

    what do you think about that one?

    for the 3rd one i have the same problem as the first one. If I drop the x^3 down so its S x^3 (1/(x^2 -1)) dx I don't know what do do.

    same problem for the 4th one.

    and for the last one, varifying that the integral is correct.
    S e^(a*u) du would = (1/a)*e^(a*u)
    but i dont know what to do with the first u in: S u*e^(a*u) du

    please help :)
  5. Aug 8, 2007 #4
    It does not work when [tex]a=0[/tex].

    But if [tex]a\not = 0[/tex] do the following:
    1)Use integration by parts.
    2)Then use the function composition [tex]t=au[/tex].
  6. Aug 9, 2007 #5

    Gib Z

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    Homework Helper

    For the first one, good idea. Take the factor is -2 out of the integral, and you know that is (arcsin x +C). Now times by -2, and we get -2 arcsin x + C.

    For the second one, I Would rewrite the numerator as 2x+6 -1, then split the integral so it looks like:

    [tex]\int \frac{2x+6}{x^2+6x -3} dx - \int \frac{1}{(x^2+3)^2 - 12} dx[/tex]. First integral is simple, let u = the denominator. Second one abit trickier, needs a trig substitution, try finding which one by yourself.

    Third one, rewrite integrand as x + x/(x^2-1), easy from there.

    Fourth one, u= x^2.

    Last one you wanted checked, finding the derivative of the RHS produced the integrand, so it is correct.
  7. Aug 9, 2007 #6


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    Do you know variable substitution. In that case, try [itex]x = \sin u[/itex] and use the identity [itex]\cos^2 u + \sin^2 u = 1[/itex]. It will also solve the first one (if the integration interval is finite, that is).

    For the second, use GibZ trick above. Note that it relies on the fact that you can split the integral in two terms, on of which has a fraction with a function [itex]f(x)[/itex] and its derivative [itex]f'(x)[/itex] on both sides of the divison bar.
  8. Aug 9, 2007 #7
    So the first one is -2 arcsin x + C. Is arcsin(x) that same as the inverse sin, written as [tex]sin^-^1(x)[/tex] ?? then the answer is [tex]-2sin^-^1(x)+C[/tex] ???

    For the second one, is the work I first tried to do wrong?
    and when you say split the integral so it looks like: [tex]\int \frac{2x+6}{x^2+6x -3} dx - \int \frac{1}{(x^2+3)^2 - 12} dx[/tex]
    Why does the denominator change? why doesn't it become: ???
    [tex]\int \frac{2x+6}{x^2+6x -3} dx - \int \frac{1}{x^2+6x -3} dx[/tex]
    ok, but to go along with your advise, the first of the two split integrals becomes just [tex]{x^2+6x -3+C}[/tex] right??? and I have no idea about the other one. please someone explain.

    Ok, for the third one, if I rewrite it as [tex]\int x+ \frac{x}{x^2-1} dx[/tex],
    then is the answer [tex]\frac{x^2}{2}+ \frac{1}{2}ln(x^2-1) dx[/tex] ???

    And for the forth one: [tex]\int \frac{x}{1+x^4}dx[/tex], if [tex]u=x^2[/tex], I dont understand what to do. please help.

    and the last one, I need to prove that this integral below is correct by showing work and evaluating the intagral to get the answer. How do I do that?
    [tex]\int ue^a^u du = \frac{e^a^u}{a}(u-\frac{1}{a})+C[/tex]
  9. Aug 10, 2007 #8

    Gib Z

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    No, for the second question: THe first integral actually becomes ln( x^2 + 6x -3). For the second part, the denominator you propose, and the way i wrote it, are the same. CHeck it out.

    Third one seems correct.

    Fourth one, if you let u= x^2, it becomes [tex] \frac{1}{2} \int \frac{1}{1+u^2} du[/tex] which is arctan u (also inverse tan u).

    Last one, integrate by parts.
  10. Aug 10, 2007 #9


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    yes, if by inverse sin you mean the function [itex]\arcsin[/itex] such that [itex]\sin\left( \arcsin(x) \right) = x[/itex]
    (and not 1/sin) :smile:.
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