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Math Help - Two Integrals

  1. #1
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    Two Integrals

    1) x^3*(x^2-1)^(1/3) dx

    2) 1/(xln(x^3)) dx

    Appreciate all help given. Thank you.
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  2. #2
    Super Member PaulRS's Avatar
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    2. Let u=\ln(x) and remember that: \ln(x^3)=3\cdot{\ln(x)}
    1. Note that: <br />
x^3  \cdot \left( {x^2  - 1} \right)^{\tfrac{1}<br />
{3}}  = x \cdot \left( {x^2  - 1 + 1} \right) \cdot \left( {x^2  - 1} \right)^{\tfrac{1}<br />
{3}}  = x \cdot \left( {x^2  - 1} \right)^{\tfrac{1}<br />
{3} + 1}  + x \cdot \left( {x^2  - 1} \right)^{\tfrac{1}<br />
{3}} <br />
    and now try a substitution
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by Hibijibi View Post
    1) x^3*(x^2-1)^(1/3) dx
    =x^2 \cdot x \cdot (x^2-1)^{\frac 13}

    Integrate by parts :

    u(x)=x^2

    \begin{aligned} v'(x) &=x \cdot (x^2-1)^{\frac 13} \\<br />
&=\frac 12 \cdot 2x \cdot (x^2-1)^{\frac 13} \end{aligned}

    \Longrightarrow \begin{aligned} v(x) &=\frac 12 \cdot \frac{(x^2-1)^{\frac 43}}{\frac 43} \\<br />
&=\frac 38 \cdot (x^2-1)^{\frac 43} \end{aligned}

    etc...
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  4. #4
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    So I rewrite the ln function as 1/ (x (3*ln(x))) and use u to equal lnx, and u'= 1/x, and then what?

    Edit: i got the function down to 1/3 intergral of 1/u du, and how do I get past this?
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  5. #5
    Moo
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    Quote Originally Posted by Hibijibi View Post
    So I rewrite the ln function as 1/ (x (3*ln(x))) and use u to equal lnx, and u'= 1/x, and then what?
    u=\ln x

    -> du=\frac{dx}{x}

    dx=x du

    do the substitutions
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  6. #6
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    Thanks, I got it. Except I I used du = 1/x dx, since 1/x dx is in the function already and I solved. Appreciate it!
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  7. #7
    Moo
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    Quote Originally Posted by Hibijibi View Post
    Thanks, I got it. Except I I used du = 1/x dx, since 1/x dx is in the function already and I solved. Appreciate it!
    I'm sorry about it... Actually I'm a bit lost concerning the way people learn how to do substitutions
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  8. #8
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    Haha everyone to their own right? Thank you very much Moo! *hugs*
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