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nubeer Group Title

integrate (1/(x^2+8x +25 )^0.5)

  • 2 years ago
  • 2 years ago

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  1. Andras Group Title
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    This is a hard one x^2+8x+25=(x+4)^2+9 so use sub u=x+4 du=dx New integral: 1/(u^2+9)^0.5

    • 2 years ago
  2. nubeer Group Title
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    hmmm ya i also reached at this point but what after this?

    • 2 years ago
  3. Andras Group Title
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    thinking of the next substitution, maybe some trig function

    • 2 years ago
  4. nubeer Group Title
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    hmm yes i think that would be tan but how and why

    • 2 years ago
  5. Andras Group Title
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    I know! Recall that tan^2(x)+1=sec^2(x)

    • 2 years ago
  6. Andras Group Title
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    so we need to make 9tan^2(x) from u to get 9(tan^2(x)+1)

    • 2 years ago
  7. Andras Group Title
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    let u=3tan(z) du=3sec^2(z) dz

    • 2 years ago
  8. Andras Group Title
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    So the integral will be: 3sec^2(z)/3sec(z)=sec(z)

    • 2 years ago
  9. Andras Group Title
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    What the integral of secz? log(tanz+secz) (just looked it up)

    • 2 years ago
  10. Andras Group Title
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    we have u=3tanz expressing z=tan^-1(u/3)

    • 2 years ago
  11. nubeer Group Title
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    thanks man.. well pretty much hit the answer but can u explain this 1 step let u=3tan(z) du=3sec^2(z) dz.. why we suppose 3 tan(z) why not just tan

    • 2 years ago
  12. Andras Group Title
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    because if it is 3tan than (3tan)^2=9tan^2(x) so we can factor out 9 to get 9(tan^2(x)+1)

    • 2 years ago
  13. nubeer Group Title
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    ohh so just that was the reason to put 3?

    • 2 years ago
  14. Andras Group Title
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    yes

    • 2 years ago
  15. Andras Group Title
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    Hope it helped, now you just need to plug back the z and u to get x

    • 2 years ago
  16. FoolForMath Group Title
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    do you know about hyperbolic functions ?

    • 2 years ago
  17. FoolForMath Group Title
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    sinhx ?

    • 2 years ago
  18. FoolForMath Group Title
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    and this is a elementary problem no need to think hard on this one

    • 2 years ago
  19. Andras Group Title
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    e^x+e^-x/2 how would that help?

    • 2 years ago
  20. Zarkon Group Title
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    \[\frac{d}{dx}\operatorname{arsinh}(x) =\frac{1}{\sqrt{x^{2}+1}}\]

    • 2 years ago
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