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Jonask

  • 3 years ago

nth integral

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  1. Jonask
    • 3 years ago
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    Let \[I_n=\huge\int \frac{x^n}{\sqrt{x^2+a}}\] \[n \in N,a>0,and \space I_n=(a,\infty)\]

  2. Jonask
    • 3 years ago
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    Find \[I_0,I_1 ,I_n\]

  3. Jonask
    • 3 years ago
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    I did the first two but three i dont understand

  4. Jonask
    • 3 years ago
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    actually it says show that \[I_n=\frac{1}{n}x^{n-1}\sqrt{x^2+a}-\frac{n-1}{n}aI_{n-2}\]

  5. Jonask
    • 3 years ago
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    \[I_0=\ln(x+\sqrt{x^2+a})-\ln\sqrt{a}+c,I_1=\frac{1}{2}\sqrt{x^2+a}+c\]

  6. hartnn
    • 3 years ago
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    tried integrating by parts ?

  7. Jonask
    • 3 years ago
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    yes\[ f=x^n,f'=nx^{n-1}\] \[g'=\frac{1}{\sqrt{x^2+a}}.g=I_0\] \[I_n=x^n(I_0)-\int I_0nx^{n-1}\]

  8. Jonask
    • 3 years ago
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    \[I_0\] is known above

  9. Jonask
    • 3 years ago
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    \[I_n=x^n\ln(x+\sqrt{x^2+a})-n\int x^{n-1}\ln(x+\sqrt{x^2+a})\]

  10. Jonask
    • 3 years ago
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    using parts twice \[\int(x^{n-1})\ln(x+\sqrt{x^2+a)}\] \[f=x^n-1,f'=(n-1)x^{n-2}\] \[g'=\ln(x+\sqrt{x^2+a)}\] not sure wat to use for g

  11. Jonask
    • 3 years ago
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    sry \[f=x^{n-1}\]

  12. hartnn
    • 3 years ago
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    let me try lil bit different, \(I_n=\huge\int \frac{x^n}{\sqrt{x^2+a}}dx=\huge\int \frac{x.x^{n-1}}{\sqrt{x^2+a}}dx\) now let f(x) = x^{n-1} , g'(x) =x/(...)

  13. hartnn
    • 3 years ago
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    just a thought, donno whether it'll be useful

  14. Jonask
    • 3 years ago
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    thanks it works very well,i tried it,just too lazy to post

  15. hartnn
    • 3 years ago
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    oh, it works...good to hear :)

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