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Math Help - Indefinite integrals

  1. #1
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    Exclamation Indefinite integrals

    Question:
     \int \sqrt[3] {3x+5}.dx

    So the equation will be  \int (3x+5)^\frac{1}{3}.dx

     u=3x+5 \rightarrow \int u^\frac{1}{3}
     du=3
    After this step i dont know what to do , do i divide by 3 to produce 1?
    Please answer in step .Thank you
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int \sqrt[3] {3x+5}.dx

    So the equation will be  \int (3x+5)^\frac{1}{3}.dx

     u=3x+5 \rightarrow \int u^\frac{1}{3}
     du=3
    Here's your problem! du is NOT equal to 3, du= 3dx.
    And from that it should be easy to see that dx= (1/3)du.

    After this step i dont know what to do , do i divide by 3 to produce 1?
    Please answer in step .Thank you
    I have no idea what you could mean by "do i divide by 3 to produce 1?". Divide what by 3? "Produce 1" where and how? The more precise you are in mathematics, the bettter. Practise being precise.
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  3. #3
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int \sqrt[3] {3x+5}.dx

    So the equation will be  \int (3x+5)^\frac{1}{3}.dx

     u=3x+5 \rightarrow \int u^\frac{1}{3}
     du=3
    After this step i dont know what to do , do i divide by 3 to produce 1?
    Please answer in step .Thank you
    u=3x+5 then

    du=3dx

    \frac{1}{3}du=dx.

    So, we have

    \int(3x+5)^{1/3}dx=\int{u^{1/3}}\frac{1}{3}du
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  4. #4
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    Exclamation

    Quote Originally Posted by HallsofIvy View Post
    Here's your problem! du is NOT equal to 3, du= 3dx.
    And from that it should be easy to see that dx= (1/3)du.


    I have no idea what you could mean by "do i divide by 3 to produce 1?". Divide what by 3? "Produce 1" where and how? The more precise you are in mathematics, the bettter. Practise being precise.
    Answer:
     \int \frac{1}{4} (3x+5)^\frac{4}{3} +C

    Is my answer Correct ?
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  5. #5
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int \sqrt[3] {3x+5}.dx

    So the equation will be  \int (3x+5)^\frac{1}{3}.dx

     u=3x+5 \rightarrow \int u^\frac{1}{3}
     du=3
    After this step i dont know what to do , do i divide by 3 to produce 1?
    Please answer in step .Thank you
    It might help to write it out like this

    u = 3x + 5 so \frac{du}{dx} = 3.


    Now work on the integral

    \int{(3x + 5)^{\frac{1}{3}}\,dx} = \frac{1}{3}\int{(3x + 5)^{\frac{1}{3}} \cdot 3\,dx}

     = \frac{1}{3}\int{u^{\frac{1}{3}}\,\frac{du}{dx}\,dx  }

     = \frac{1}{3}\int{u^{\frac{1}{3}}\,du}.


    Might be an extra step but easier to avoid making mistakes...
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  6. #6
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    Question

    Quote Originally Posted by Prove It View Post
    It might help to write it out like this

    u = 3x + 5 so \frac{du}{dx} = 3.


    Now work on the integral

    \int{(3x + 5)^{\frac{1}{3}}\,dx} = \frac{1}{3}\int{(3x + 5)^{\frac{1}{3}} \cdot 3\,dx}

     = \frac{1}{3}\int{u^{\frac{1}{3}}\,\frac{du}{dx}\,dx  }

     = \frac{1}{3}\int{u^{\frac{1}{3}}\,du}.




    Might be an extra step but easier to avoid making mistakes...

    Answer:
     \frac{1}{2} (3x+5)^\frac{4}{3} +C

    Is my answer Correct ?
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  7. #7
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    No.

    \frac{1}{3}\int{u^{\frac{1}{3}}\,du} = \frac{1}{3}\cdot\frac{3}{4}u^{\frac{4}{3}} + C

     = \frac{1}{4}u^{\frac{4}{3}} + C

     = \frac{1}{4}(3x + 5)^{\frac{4}{3}} + C.
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  8. #8
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    Quote Originally Posted by mj.alawami View Post
    Answer:
     \int \frac{1}{4} (3x+5)^\frac{4}{3} +C

    Is my answer Correct ?
    What do you get if you differentiate that? Did you even try that?
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  9. #9
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    substitution isn't needed in this question.
    If you remember this:

    if you have a derivative that is in the form: (ax+b)^n
    then the integral will be: [(ax+b)^n+1] / a(n+1)

    so just substitute your question into that rule.
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  10. #10
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    Quote Originally Posted by bhuang View Post
    substitution isn't needed in this question.
    If you remember this:

    if you have a derivative that is in the form: (ax+b)^n
    then the integral will be: [(ax+b)^n+1] / a(n+1)

    so just substitute your question into that rule.
    Don't try remembering lots of rules, it gets to be too much of a headache.

    It's easier to remember a few basic rules and to use substitution for the rest.

    Besides, the rule you speak of comes from substitution anyway.
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