Post a New Question


posted by .

Find the integral of x/ (x^4+x^2+1) from 2 to 3.

I was going to integrate using the method for partial fractions but I can't factor x^4+x^2+1.

  • Calculus -

    integrate x/(x^4+x^2+1)
    | = integrate symbol
    u = x^2
    du = 2x dx
    1/2 du = x dx

    1/2 | du/(u^2 + u + 1)

    Complete the square
    u^2 + u = -1
    u^2 + u + 1/4 = -1 + 1/4
    (u + 1/2)^2 + 3/4

    1/2 | du/((u + 1/2)^2 + 3/4)

    w = u + 1/2
    dw = du
    1/2 | dw/(w^2 + 3/4)
    1/2 | dw/(3/4 + w^2)

    And, since,
    | dx/(a^2 + x^2)= 1/a arctan x/a + C

    the integration can easily be done now.

    Can you take it from here?

  • Calculus -

    Okay, so I ended up with the expression:


    Can you confirm this is correct before I use the fundamental theorem of calculus to determine the definite integral please.

  • Calculus -


    I distributed the 1/2, but backed it out for my above answer...I don't think I made any errors backing it out.

    Here is the answer, with the 1/2 distributed. I know this answer is right because I double check it with an online integration calculator.

    1/(sqrt3)arctan (2x^2 + 1)/(sqrt3)

    I didn't rationalize the denominators, because the online answer didn't.

    Do you want me to post the final answer for you to check the definite integral?

  • Calculus -

    That would be great if you could!
    I'm working on getting the correct indefinite as you have above, but when I get it it would be great to have something to check my definite integral with :)

  • Calculus -

    1/(sqrt3)arctan(5/(29(sqrt3))= 0.0572826

    Check you values for 1/a and x/a
    maybe that's where you got hung up
    | dx/(a^2 + x^2)

    (3/4 + w^2)
    a^2 = 3/4
    a = (sqrt3)/2
    x^2 = w^2
    x = w

    So, you should have got,
    1/2(1/(sqrt3/2)arctan (w/(sqrt3/2))
    which simplified is,

    then I distributed the 1/2

    then substitute back in for u and x

    Good Luck!!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question