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Substitution rule

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    here is the answer


    3. The attempt at a solution

    My book doesn't do a good job of explaining the substitution rule. here is their explanation:


    using the solution manual and looking at how they got the answer to other questions, i've written down my own method in english that i can understand. the method has worked for two other problems, but it broke down with the above problem.

    i still don't understand what integrate with respect to u means.

    here's my method:

    1. substitute one part of the integral with u, find the derivative of that, z,
    2. multiply the whole integral by the reciprocal of z, so that z equals 1
    3. find the antiderivative of the remaining integral
    4. replace u by g(x) in the result

    the derivative of g(x) is

    step 1. 4y^3 + 8y

    multiply 12 by y^3 + 2y

    step 2. 12y^3 + 2y

    multiply the reciprocal of 1 with the result of step 2

    step 3. (12y^3 + 24y)/(4y^3 + 8y)

    simplify step 3

    step 4. 3 + 3 = 6

    we now have

    step 5. 6u^2

    take the antiderivative of 6u^2

    step 6. (6u^3)/3


    step 7 2u^3

    plug in g(x) into u

    step 8. 2(y^4 + 4y^2 + 1)^3

    The book says that the answer is

    (y^4 + 4y^2 + 1)^3
  2. jcsd
  3. Feb 2, 2012 #2
    I don't understand what you're doing from step 2 to step 4, why are you multiplying by the reciprocal?

    Can you explain what you're doing at that step and why? Where did the 3+3 from step 4 come from?
  4. Feb 2, 2012 #3
    That was the method I came up with to get the other answers and it worked.

    It looks like the book just eliminates (y^3 + 2y) why?

    I don't understand what integrate with respect to u means.

    If you just eliminate y^3 + 2y, then add 4 + 8, which is 12 and multiply the reciprocal of that to 12, you will get the answer, but I would like to know why.
  5. Feb 2, 2012 #4


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    Your integral is [itex]12\int (y^4+ 4y^2+ 1)(y^3+ 2y)dy[/itex] and you want to use the substitution [itex]u= y^4+ 4y^2+ 1[/itex]. By differentiating, [itex]du= (4y^2+ 8y)dy[/itex] or [itex]du= 4(y^2+ 2y)dy[/itex] which is the same thing as [itex](1/4)du= (y^2+ 2y)dy[/itex]. So you replace the "[itex]y^4+ 4y^2+ 1[/itex]" in the integral with u and the "[itex](y^2+ 2y)dy[/itex]" with [itex](1/4)du[/itex]

    That gives [itex]12\int u (1/4)du= 3\int dudu[/itex]

    That's all it is, just replace the formulas in x with the things in u that are equal to them.
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