Is $int frac{1}{x} dx = int frac{dx}{x}$?












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Is the integral of $(1/x) dx$ the same as the integral of $dx/x$ ?






calculus integration







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  • $begingroup$
    What did you try ?
    $endgroup$
    – Felix Marin
    Oct 17 '13 at 9:22
















3












$begingroup$


Is the integral of $(1/x) dx$ the same as the integral of $dx/x$ ?






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    What did you try ?
    $endgroup$
    – Felix Marin
    Oct 17 '13 at 9:22














3












3








3


2



$begingroup$


Is the integral of $(1/x) dx$ the same as the integral of $dx/x$ ?






calculus integration







share|cite|improve this question











$endgroup$




Is the integral of $(1/x) dx$ the same as the integral of $dx/x$ ?






calculus integration




calculus integration






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edited Jun 8 '12 at 14:24









davidlowryduda♦

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asked Dec 26 '10 at 22:17









oneatoneat

1425




1425












  • $begingroup$
    What did you try ?
    $endgroup$
    – Felix Marin
    Oct 17 '13 at 9:22


















  • $begingroup$
    What did you try ?
    $endgroup$
    – Felix Marin
    Oct 17 '13 at 9:22
















$begingroup$
What did you try ?
$endgroup$
– Felix Marin
Oct 17 '13 at 9:22




$begingroup$
What did you try ?
$endgroup$
– Felix Marin
Oct 17 '13 at 9:22










3 Answers
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$begingroup$

Yes, $int frac{dx}{x}$ is simply shorthand for $int frac1x , dx$; they mean precisely the same thing.






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    5












    $begingroup$

    Yes. $int frac{1}{x} mathrm dx = int frac{mathrm dx}{x}$.






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      0












      $begingroup$

      In $ int frac{dx}{x} $, the coefficient of the numerator inside the integral, $ dx $, when there is none, is understood to be 1.



      So yes, $ int frac{1}{x} dx $ = $ int frac{1 cdot dx}{x} $ = $ int frac{dx}{x} $.






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        3 Answers
        3






        active

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        3 Answers
        3






        active

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        active

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        active

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        9












        $begingroup$

        Yes, $int frac{dx}{x}$ is simply shorthand for $int frac1x , dx$; they mean precisely the same thing.






        share|cite|improve this answer









        $endgroup$


















          9












          $begingroup$

          Yes, $int frac{dx}{x}$ is simply shorthand for $int frac1x , dx$; they mean precisely the same thing.






          share|cite|improve this answer









          $endgroup$
















            9












            9








            9





            $begingroup$

            Yes, $int frac{dx}{x}$ is simply shorthand for $int frac1x , dx$; they mean precisely the same thing.






            share|cite|improve this answer









            $endgroup$



            Yes, $int frac{dx}{x}$ is simply shorthand for $int frac1x , dx$; they mean precisely the same thing.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 26 '10 at 22:19









            Moor XuMoor Xu

            1,84022125




            1,84022125























                5












                $begingroup$

                Yes. $int frac{1}{x} mathrm dx = int frac{mathrm dx}{x}$.






                share|cite|improve this answer









                $endgroup$


















                  5












                  $begingroup$

                  Yes. $int frac{1}{x} mathrm dx = int frac{mathrm dx}{x}$.






                  share|cite|improve this answer









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    Yes. $int frac{1}{x} mathrm dx = int frac{mathrm dx}{x}$.






                    share|cite|improve this answer









                    $endgroup$



                    Yes. $int frac{1}{x} mathrm dx = int frac{mathrm dx}{x}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 26 '10 at 22:20









                    HuyHuy

                    3,22623268




                    3,22623268























                        0












                        $begingroup$

                        In $ int frac{dx}{x} $, the coefficient of the numerator inside the integral, $ dx $, when there is none, is understood to be 1.



                        So yes, $ int frac{1}{x} dx $ = $ int frac{1 cdot dx}{x} $ = $ int frac{dx}{x} $.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          In $ int frac{dx}{x} $, the coefficient of the numerator inside the integral, $ dx $, when there is none, is understood to be 1.



                          So yes, $ int frac{1}{x} dx $ = $ int frac{1 cdot dx}{x} $ = $ int frac{dx}{x} $.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            In $ int frac{dx}{x} $, the coefficient of the numerator inside the integral, $ dx $, when there is none, is understood to be 1.



                            So yes, $ int frac{1}{x} dx $ = $ int frac{1 cdot dx}{x} $ = $ int frac{dx}{x} $.






                            share|cite|improve this answer









                            $endgroup$



                            In $ int frac{dx}{x} $, the coefficient of the numerator inside the integral, $ dx $, when there is none, is understood to be 1.



                            So yes, $ int frac{1}{x} dx $ = $ int frac{1 cdot dx}{x} $ = $ int frac{dx}{x} $.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 2 at 15:16









                            Marvin CohenMarvin Cohen

                            158117




                            158117






























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