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Basic integration

  1. Jun 14, 2017 #1
    1. The problem statement, all variables and given/known data
    Solve the following integral: [(x^2+3)/(x^8+x^6)] dx


    2. Relevant equations
    The question has also said to integrate by substitution(though other methods are welcome)
    That would mean substituting an expression in x with a variable,say, 't' such that the integral comes of the form f(t)dt ,which is eaier to evaluate using standard results,and then we can replace the value of t


    3. The attempt at a solution
    Broke down the numerator (x^2+3) into 2 terms (x^2+1) and 2 and then,separated the two fractions.
    The first term came out to be (1/x^6)dx ,which is easy to integrate,but the second expression is really problematic: [2/(x^8 + x^6)]dx
    I tried multiplying certain powers of x both to the numerator and denominator,but that hasn't worked well so far.(thought of substituting trigo terms,but that doesn't look promising either)
    So..stuck
     
  2. jcsd
  3. Jun 14, 2017 #2

    andrewkirk

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    Try working on the denominator rather than the numerator.

    Use partial fraction decomposition to express the fraction as the sum
    $$\frac{something}{x^6} + \frac{something\ else}{x^2+1}$$

    The integration should then be easy.
     
  4. Jun 14, 2017 #3
    Something=3
    Something else=-2/x^4

    The first term is easy enough,but then a similar problem is coming with the second term.
    How to integrate
    -2/(x^4+x^6) ??
     
  5. Jun 14, 2017 #4

    Ray Vickson

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    Convert properly to partial fractions:
    $$\frac{-2}{x^4+x^6} = \frac{-2}{x^4(1+x^2)} = \frac{A}{x^2} + \frac{B}{x^4}+\frac{C}{1+x^2}.$$
    Alternatively, let ##t = x^2## and convert
    $$\frac{-2}{t^2(1+t)}$$
    to partial fractions, then put back ##t = x^2## later.

    Also, you could have saved yourself a lot of trouble by doing it correctly from the start. Putting ##x^2 = t## in your ##f(x) = (3+x^2)/(x^6+x^8)## gives
    $$\frac{3+t}{t^3+t^4} = \frac{3+t}{t^3(1+t)} = \frac{A + B t + C t^2}{t^3} +\frac{D}{1+t}.$$
     
    Last edited: Jun 14, 2017
  6. Jun 15, 2017 #5
    Yup,got the asnwer.Thanks!!
     
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