How to calculate the indefinite integral of $(x-1)^{frac 12}-(x-3)$?











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Why is the following integral wrong?



$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12(x-3)^2+c $$



The answer given by my textbook is:
$$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12x^2+3x+c $$



I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.






calculus integration indefinite-integrals







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    up vote
    3
    down vote

    favorite












    Why is the following integral wrong?



    $$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12(x-3)^2+c $$



    The answer given by my textbook is:
    $$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12x^2+3x+c $$



    I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.






    calculus integration indefinite-integrals







    share|cite|improve this question


























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      Why is the following integral wrong?



      $$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12(x-3)^2+c $$



      The answer given by my textbook is:
      $$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12x^2+3x+c $$



      I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.






      calculus integration indefinite-integrals







      share|cite|improve this question















      Why is the following integral wrong?



      $$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12(x-3)^2+c $$



      The answer given by my textbook is:
      $$ int[(x-1)^{frac 12}-(x-3)]dx=frac 23 (x-1)^{frac32}-frac12x^2+3x+c $$



      I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.






      calculus integration indefinite-integrals




      calculus integration indefinite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      AccidentalFourierTransform

      1,427827




      1,427827










      asked 2 days ago









      John Arg

      266




      266






















          1 Answer
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          accepted










          Note that
          begin{align}
          frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
          end{align}
          so both answers are valid.






          share|cite|improve this answer























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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            6
            down vote



            accepted










            Note that
            begin{align}
            frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
            end{align}
            so both answers are valid.






            share|cite|improve this answer



























              up vote
              6
              down vote



              accepted










              Note that
              begin{align}
              frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
              end{align}
              so both answers are valid.






              share|cite|improve this answer

























                up vote
                6
                down vote



                accepted







                up vote
                6
                down vote



                accepted






                Note that
                begin{align}
                frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
                end{align}
                so both answers are valid.






                share|cite|improve this answer














                Note that
                begin{align}
                frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x-3)^2+c &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}(x^2-6x+9)+c\ &= frac{2}{3}(x-1)^{3/2}-frac{1}{2}x^2+3xunderbrace{-frac{9}{2}+c}_{text{constant}},
                end{align}
                so both answers are valid.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered 2 days ago









                MisterRiemann

                5,6691624




                5,6691624






























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