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exraven

  • 3 years ago

evaluating indefinite integral

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  1. exraven
    • 3 years ago
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    \[\int\limits \frac{x^4 + 1}{x^2 \sqrt{x^4 - 1}}dx\]

  2. UnkleRhaukus
    • 3 years ago
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    \[u=x^4-1\] \[(u+1)^{1/4}=x\] \[\frac 14(u+1)^{-3/4}\text du=\text dx\] \[\int\limits \frac{u+2}{(u+1)^{2/4} \sqrt{u}}\frac 14(u+1)^{-3/4}\text du\]

  3. UnkleRhaukus
    • 3 years ago
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    hmm

  4. exraven
    • 3 years ago
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    I have already tried that substitution and stuck

  5. UnkleRhaukus
    • 3 years ago
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    maybe there is a trig substitution

  6. Mimi_x3
    • 3 years ago
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    \[\int\limits\frac{(x^{2})^{2}+1}{x^{2}\sqrt{(x^{2})^{2}-1}} \] maybe a trig sub..u^2 = tan\theta or u^2 = sin\theta

  7. Mimi_x3
    • 3 years ago
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    i mean x^2 = tan\theta or x^2 = sin\theta

  8. mukushla
    • 3 years ago
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    this is guessing game...maybe\[x^2=\sec u\]

  9. Mimi_x3
    • 3 years ago
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    yeah my bad..but i think x^2 = tan(u) would work as well..

  10. exraven
    • 3 years ago
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    here it is\[\int\limits \frac{\sec^2 u + 1}{2 \sqrt{\sec u}} du\]

  11. exraven
    • 3 years ago
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    I don't know what to do with the sqrt{sec u}

  12. Mimi_x3
    • 3 years ago
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    secx = 1/cosx maybe weirstrass substitution?

  13. Mimi_x3
    • 3 years ago
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    wait..wont work http://www.wolframalpha.com/input/?i=integrate+sqrt%28secx%29

  14. exraven
    • 3 years ago
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    there must be a tricky substitution

  15. cruffo
    • 3 years ago
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    Have you tried splitting the integral up like this: ?

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  16. exraven
    • 3 years ago
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    it won't work... http://www.wolframalpha.com/input/?i=integrate+%28x^2%29%2F%28sqrt%28x^4+-1%29%29+dx

  17. Mimi_x3
    • 3 years ago
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    then what about this \[\int\limits\frac{(secu)^{2}+1}{(secu)^{2}\sqrt{(secu)^{2}-1}} *\frac{tanusecu}{2\sqrt{secu}} \]? its not going to look nice tho

  18. exraven
    • 3 years ago
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    \[\int\limits \frac{\sec^2 u+1}{2\sqrt{\sec u}} du\]

  19. Mimi_x3
    • 3 years ago
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    hmm..this is hard..would integration by parts work?

  20. exraven
    • 3 years ago
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    haven't tried that, but I doubt it will work

  21. amistre64
    • 3 years ago
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    \[sec=\sqrt{tan^2+1}\]\[sec^2=tan^2+1\] \[\int\limits \frac{\tan^2 u+2}{2tan^2u+2} du\] or am i missing something?

  22. mukushla
    • 3 years ago
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    my last hope is x=e^t

  23. Mimi_x3
    • 3 years ago
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    x=e^t? i might sound stupid; but where did e^t come from?

  24. amistre64
    • 3 years ago
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    i missed something :)\[x^{1/4}\ne x\]

  25. mukushla
    • 3 years ago
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    well maybe that will lead us to some sinh and cosh

  26. mukushla
    • 3 years ago
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    yeah x=e^t works

  27. mukushla
    • 3 years ago
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    x=e^t dx=e^t dt\[\int \frac{e^{4t} + 1}{e^{2t} \sqrt{e^{4t} - 1}}e^t\text{d}t=\int \frac{e^{2t} + e^{-2t}}{ \sqrt{e^{2t} - e^{-2t}}}\text{d}t=\int \frac{2\cosh t}{\sqrt{\sinh t}} \text{d}t\]

  28. mukushla
    • 3 years ago
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    OOoopS typo again\[\int \frac{2\cosh 2t}{\sqrt{2\sinh 2t}} \text{d}t\]

  29. exraven
    • 3 years ago
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    nice, it works!

  30. mukushla
    • 3 years ago
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    :)

  31. exraven
    • 3 years ago
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    now I can rest peacefully lol

  32. exraven
    • 3 years ago
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    ty all for helping :)

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