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    I did a question and got an answer which I believe is correct as I got a follow on part right, however the book answer seems to have been simplified and I don't know how.

    Question:
    Using the substituion t^2=x+1 find \int \frac{x}{\sqrt(x+1)}\ dx

    My answer:
    2\frac{(x+1)^{3/2}}{3}-2(x+1)^{1/2}+c

    Books answer (which wolfram agrees with too):
    \frac{2}{3}(x-2)\sqrt(x+1)+c

    Could somebody please check my answer and then explain how this simplifies to the one in the book, thankyou
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    (Original post by fayled)
    I did a question and got an answer which I believe is correct as I got a follow on part right, however the book answer seems to have been simplified and I don't know how.

    Question:
    Using the substituion t^2=x+1 find \int \frac{x}{\sqrt(x+1)}\ dx

    My answer:
    2\frac{(x+1)^{3/2}}{3}-2(x+1)^{1/2}+c

    Books answer (which wolfram agrees with too):
    \frac{2}{3}(x-2)\sqrt(x+1)+c

    Could somebody please check my answer and then explain how this simplifies to the one in the book, thankyou
    you have the same answer

    Take out the common factor of \sqrt{x+1}
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    (Original post by fayled)
    I did a question and got an answer which I believe is correct as I got a follow on part right, however the book answer seems to have been simplified and I don't know how.

    Question:
    Using the substituion t^2=x+1 find \int \frac{x}{\sqrt(x+1)}\ dx

    My answer:
    2\frac{(x+1)^{3/2}}{3}-2(x+1)^{1/2}+c

    Books answer (which wolfram agrees with too):
    \frac{2}{3}(x-2)\sqrt(x+1)+c

    Could somebody please check my answer and then explain how this simplifies to the one in the book, thankyou
    They're the same - your first term is (2/3)(x+1)\sqrt{x+1} so take out a common factor of \sqrt{x+1} and rearrange the factor multiplying it.
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    (Original post by TenOfThem)
    you have the same answer

    Take out the common factor of \sqrt{x+1}
    Oh why didn't I think of that. Thanks.
 
 
 
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