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Homework Help: Integration using tables

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int x^{5}\sqrt{x^{2}+1}dx[/tex]

    2. Relevant equations

    Using various tables for integration.

    3. The attempt at a solution

    So I want to manipulate it, perhaps get it to look like [itex]\sqrt{a^{2}+u^{2}}[/itex], [itex]\sqrt{a^{2}-u^{2}}[/itex], [itex]\sqrt{u^{2}-a^{2}}[/itex], or maybe even [itex]\sqrt{2au-u^{2}}[/itex]. So I attempt to do some substitutions:

    if i let u = x^2, then du = 2x dx and x = sqrt(u)

    so i rearrange things and get:

    [tex]\frac{1}{2}\int u^{2}\sqrt{u+1}du[/tex]

    and while i may (or may not) be able to do this, the question is asking that i use the tables in the back of my textbook, and none of them are given in this form...so...

    if i let u = x^4, then in the end, x will equal the cube root of u, which doesn't help at all.

    I tried u = x^3, u = x^5, and u = x^2+1 and had no luck...I used wolfram alpha and even it couldn't give me any steps that would help guide me in the right direction.

    how do i manipulate this integrand to get something in one of the forms i gave earlier, [itex]\sqrt{a^{2}+u^{2}}[/itex], [itex]\sqrt{a^{2}-u^{2}}[/itex], [itex]\sqrt{u^{2}-a^{2}}[/itex], or [itex]\sqrt{2au-u^{2}}[/itex]. If you can think of any others please let me know.

    If not, I guess I just need to let u = x^2 and then substitute. Then after that, we have
    [tex]\frac{1}{2}\int u^{2}\sqrt{u+1}du[/tex]

    then i guess do integration by parts? even then I'll have a [tex]\int \frac{u^{3}}{\sqrt{u+1}}du[/tex], which isn't part of any tables...

    i think i am probably complicating things, but i'm not seeing what technique to use, given that we have to use some sort of table..

    thank you in advance for any help!
    Last edited: Jun 10, 2010
  2. jcsd
  3. Jun 10, 2010 #2
    its easier to use trigonometric substitution
  4. Jun 10, 2010 #3
    i tried that, let x = tan u and dx = sec^2u du. Then use substitution again, let t = tan u and again dt = sec^2t dt, but through ALL that, you'll have an extra factor of sec u that is still in terms of you, but you have changed the variable of integration to t. I've been having a hard time getting rid of that extra factor.

    any other suggestions?
  5. Jun 10, 2010 #4


    Staff: Mentor

    Try integration by parts, with u = x4 and dv = x*sqrt(x2 + 1)dx

    That gets you to an integral whose integrand is x3 * sqrt(x2 + 1). Use integration by parts again.
  6. Jun 10, 2010 #5
    no, the integrand you are talking about is not x^3 * sqrt(x^2 + 1), the integrand is x^3 * (x^2+1)^(3/2), and that doesn't get me very far.
  7. Jun 10, 2010 #6


    Staff: Mentor

    I guess the moral here is be careful doing integration by parts in your head.

    Since that advice didn't help, I would go with trig substitution, like annoymage advised.

    Using x = tan u, the original integral becomes
    [itex]\int tan^5 u~sec u~sec^2 u~du[/itex]

    With a little manipulation, this becomes
    [itex]\int tan^4 u~sec^2 u~sec u~tan u du[/itex]

    Convert all the tan factors to sec factors using the identity tan2u = sec2u - 1, and you should get an integral that you can use an ordinary substitution on.

    Hint: don't mess with the sec u tan u part.
  8. Jun 10, 2010 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    After your first substitution, IMO it's pretty clear that v=u+1 leads to something that can be handled easily.

    (note that this implies u=x^2+1 would have worked out too....)

    Are you sure none of the tables you have access to have an integrand that consists of a power of x multiplied by the square root of something?
  9. Jun 10, 2010 #8
    after the first substitution v = u+1 or when I used u = x^2+1, i believe the integrand was still fairly difficult...and the tables we were given don't include any power of x higher than 2, multiplied by the square root of something.

    my instructor said to use integration by parts twice, but he also noted that trig substitution and using tables would have been rather difficult.
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