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Math Help - anti-derivative

  1. #1
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    anti-derivative

    i tried to solve this problem using substitution (as per someones suggestion) but the answer my prof came up with differs from mine.

    he got \frac{4}{15}(x^3+3x+2)^\frac{5}{4} + C

    here's the problem:

    find the antiderivative

    f(x) = (x^2+1)\sqrt[4]{x^3+3x+2}

    i used substitution

    u = x^3 + 3x + 2
    h(u) = 3x^2 + 3

    then did:
    <br />
\frac{1}{3}*3(x^2+1)\sqrt[4]{x^3+3x+2}

    which gave me:
    <br />
\frac{1}{3}\sqrt[4]{u} du

    and then:
    <br />
\frac{1}{3}\sqrt[4]{x^3+3x+2} + c


    what am i doing wrong, and how do i get the answer he got?
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by paguy30 View Post
    i tried to solve this problem using substitution (as per someones suggestion) but the answer my prof came up with differs from mine.

    he got \frac{4}{15}(x^3+3x+2)^\frac{5}{4} + C

    here's the problem:

    find the antiderivative

    f(x) = (x^2+1)\sqrt[4]{x^3+3x+2}

    i used substitution

    u = x^3 + 3x + 2
    h(u) = 3x^2 + 3

    then did:
    <br />
\frac{1}{3}*3(x^2+1)\sqrt[4]{x^3+3x+2}

    which gave me:
    <br />
\frac{1}{3}\sqrt[4]{u} du

    and then:
    <br />
\frac{1}{3}\sqrt[4]{x^3+3x+2} + c


    what am i doing wrong, and how do i get the answer he got?
    let u = x^3 + 3x + 2

    du = (3x^{2}+3) dx = 3(x^{2}+1) dx \implies dx = \frac{du}{3(x^{2}+1)}

    and then your integral becomes:

    \int (x^{2}+1) \times u^{\frac{1}{4}} \frac{du}{3(x^2+1)} = \frac{1}{3} \int u^{\frac{1}{4}} du

    now integrate, ans then substitute u
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  3. #3
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    ah awesome thanks so much.
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