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Homework Help: Integration by substituition question.

  1. Jan 14, 2005 #1
    I'm able to do this question but my answer is different from that in the book.

    [tex] \int \frac {x^3}{\sqrt{ x^2 -1}} dx [/tex]

    what I did was to take the substituition [tex] u= (x^2-1)^1^/^2 [/tex]

    so, [tex] x^2 -1 = u^2 [/tex]

    [tex]x^2 = u^2+1 [/tex]
    [tex] x= (u^2+1)^1^/^2[/tex]
    [tex] x^3= (u^2+1)^3^/^2 [/tex]
    [tex] dx= \frac {1}{2}(u^2+1)^-^1^/^2 (2u)du [/tex]
    [tex] dx= u(u^2+1)^-^1^/^2 du [/tex]

    [tex] \int \frac {x^3}{\sqrt{ x^2 -1}} dx= \int \frac {(u^2+1)^3^/^2}{u} . \frac {u}{(u^2+1)^1^/^2} du [/tex]

    which simplifies to,

    [tex] \int u^2+1 du [/tex]
    [tex] \frac {1}{3} u^3 +u+C [/tex]
    [tex] \frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2 [/tex]

    that's my final answer but the book gave,

    [tex] \frac {1}{3} (x^2+2)\sqrt{x^2-1}+C [/tex]

    where does my mistake lie?
  2. jcsd
  3. Jan 14, 2005 #2


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    Are you sure your answer is not the book's answer?
  4. Jan 14, 2005 #3


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    Yap,it's the same "animal".It's just the fur is a little shady...

  5. Jan 14, 2005 #4
    if your ever unsure of your answer ...you can always take the derivative of your answer and see if you get the integrand.
  6. Jan 14, 2005 #5
    your answer is fine... you need one more step to get the text answer
    [tex] \frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2 [/tex]
    [tex] = (x^2-1)^1^/^2 (\frac {1}{3}(x^2-1) +1)[/tex]
    [tex] = (x^2-1)^1^/^2 (\frac {1}{3}x^2-\frac {1}{3}+1)[/tex]
    [tex] = (x^2-1)^1^/^2 (\frac {1}{3}x^2+\frac {2}{3})[/tex]
    [tex] \frac {1}{3} (x^2+2)\sqrt{x^2-1}[/tex] :surprised:
  7. Jan 14, 2005 #6


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    Isn't y^(3/2) = (y)(y^(1/2)) ? Rearrange and simplify what you've got and it should come out the same.

    EDIT : NM, Vincent has shown the working
  8. Jan 15, 2005 #7
    hey thanks, that's a very good tip !

    I am wondering if the answer which I gave is in a different form from the one in the answer script during an exam, will I be penalised?
  9. Jan 15, 2005 #8


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    Unless your teacher is a narrow minded s.o.b.,i don't see why.If i were u,on this integral i would have gone for another substitution,using hyperbolic sine and cosine.

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