Integral utter confusion with substition and dx/du












0












$begingroup$


I need to find the indefinite integral I = $$int e^x (1+e^x)^{frac{1}{2}}$$



by using a proper substition method. I tried it on https://www.integral-calculator.com and it gave the following explanation



Screenshot of integral-calculator



Now this is where it gets confusing, when substituting $u = (1+e^x)$ I would suspect we got $$int e^x sqrt{(1+e^x)} = int e^xsqrt{u} $$
But that's not what the screenshot says.



It says that the original problem equals $int sqrt{u} du$ and also that $dx = e^{-x}$. Now where does this $e^{-x}$ come from? The antiderivite of $e^x$ is still $e^x$. And the antiderivative of $e^x+1$ is $e^x+x$ so where does that negative number come from and why is it used?



So those are my questions.
1. Why is $$int e^x sqrt{(1+e^x)} = int sqrt{u} du $$
2. Where does the $dx = e^{-x}$ come from
3. WHy is it even necessary?






integration indefinite-integrals substitution







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$endgroup$












  • $begingroup$
    You should not forget about writing $dx $ every time you deal with an integral.
    $endgroup$
    – user
    Jan 10 at 20:46
















0












$begingroup$


I need to find the indefinite integral I = $$int e^x (1+e^x)^{frac{1}{2}}$$



by using a proper substition method. I tried it on https://www.integral-calculator.com and it gave the following explanation



Screenshot of integral-calculator



Now this is where it gets confusing, when substituting $u = (1+e^x)$ I would suspect we got $$int e^x sqrt{(1+e^x)} = int e^xsqrt{u} $$
But that's not what the screenshot says.



It says that the original problem equals $int sqrt{u} du$ and also that $dx = e^{-x}$. Now where does this $e^{-x}$ come from? The antiderivite of $e^x$ is still $e^x$. And the antiderivative of $e^x+1$ is $e^x+x$ so where does that negative number come from and why is it used?



So those are my questions.
1. Why is $$int e^x sqrt{(1+e^x)} = int sqrt{u} du $$
2. Where does the $dx = e^{-x}$ come from
3. WHy is it even necessary?






integration indefinite-integrals substitution







share|cite|improve this question









$endgroup$












  • $begingroup$
    You should not forget about writing $dx $ every time you deal with an integral.
    $endgroup$
    – user
    Jan 10 at 20:46














0












0








0





$begingroup$


I need to find the indefinite integral I = $$int e^x (1+e^x)^{frac{1}{2}}$$



by using a proper substition method. I tried it on https://www.integral-calculator.com and it gave the following explanation



Screenshot of integral-calculator



Now this is where it gets confusing, when substituting $u = (1+e^x)$ I would suspect we got $$int e^x sqrt{(1+e^x)} = int e^xsqrt{u} $$
But that's not what the screenshot says.



It says that the original problem equals $int sqrt{u} du$ and also that $dx = e^{-x}$. Now where does this $e^{-x}$ come from? The antiderivite of $e^x$ is still $e^x$. And the antiderivative of $e^x+1$ is $e^x+x$ so where does that negative number come from and why is it used?



So those are my questions.
1. Why is $$int e^x sqrt{(1+e^x)} = int sqrt{u} du $$
2. Where does the $dx = e^{-x}$ come from
3. WHy is it even necessary?






integration indefinite-integrals substitution







share|cite|improve this question









$endgroup$




I need to find the indefinite integral I = $$int e^x (1+e^x)^{frac{1}{2}}$$



by using a proper substition method. I tried it on https://www.integral-calculator.com and it gave the following explanation



Screenshot of integral-calculator



Now this is where it gets confusing, when substituting $u = (1+e^x)$ I would suspect we got $$int e^x sqrt{(1+e^x)} = int e^xsqrt{u} $$
But that's not what the screenshot says.



It says that the original problem equals $int sqrt{u} du$ and also that $dx = e^{-x}$. Now where does this $e^{-x}$ come from? The antiderivite of $e^x$ is still $e^x$. And the antiderivative of $e^x+1$ is $e^x+x$ so where does that negative number come from and why is it used?



So those are my questions.
1. Why is $$int e^x sqrt{(1+e^x)} = int sqrt{u} du $$
2. Where does the $dx = e^{-x}$ come from
3. WHy is it even necessary?






integration indefinite-integrals substitution




integration indefinite-integrals substitution






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share|cite|improve this question











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asked Jan 10 at 20:14









Wouter VandenputteWouter Vandenputte

1295




1295












  • $begingroup$
    You should not forget about writing $dx $ every time you deal with an integral.
    $endgroup$
    – user
    Jan 10 at 20:46


















  • $begingroup$
    You should not forget about writing $dx $ every time you deal with an integral.
    $endgroup$
    – user
    Jan 10 at 20:46
















$begingroup$
You should not forget about writing $dx $ every time you deal with an integral.
$endgroup$
– user
Jan 10 at 20:46




$begingroup$
You should not forget about writing $dx $ every time you deal with an integral.
$endgroup$
– user
Jan 10 at 20:46










4 Answers
4






active

oldest

votes


















2












$begingroup$

$(1)$ Because when you make the substitution $color{purple}{u = e^x+1}$, taking the derivative of $u$ yields $frac{du}{dx} = e^x$. Hence, $color{blue}{e^xdx = frac{du}{dx}dx = du}$. The integral can therefore be rewritten:



$$int color{blue}{e^x}sqrt{color{purple}{e^x+1}} color{blue}{dx} implies intsqrt{color{purple}{u}}color{blue}{du}$$



$(2)$ In the previous part, note that we get $frac{du}{dx} = e^x$. This can be rewritten as $frac{du}{e^x} = dx$, and $e^x$ in the denominator can be written as $e^{-x}$ in the numerator, so $dx = e^{-x}du$.



$(3)$ It’s certainly not necessary. It makes the integral easier to evaluate.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $e^{x}-1 =u rightarrow e^xdx=du$ (I suggest writing it this way) so the integral after the substitution looks like this:
    $int (e^x sqrt{e^x-1})dx=int(sqrt{u})du=cfrac{2u^{3/2}}{3} + C=cfrac{2(e^x-1)^{3/2}}{3}+C$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      The way I learnt it more then sixty years ago.



      $$I=int e^x sqrt{e^x+1},dx$$
      $$e^x+1=uimplies x=log(u-1) implies dx=frac{du}{u-1}$$ making
      $$I=int (u-1), sqrt u ,frac{du}{u-1}=int sqrt u ,du=cdots$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Another approach:



        begin{equation}
        I = int e^x sqrt{e^x + 1}:dx
        end{equation}



        Here let $u^2 = e^x + 1$. Employing Implicit Differentiation we see



        begin{equation}
        frac{d}{dx}left[u^2right] = frac{d}{dx}left[e^x + 1 right] rightarrow 2u frac{du}{dx} = e^x rightarrow dx = frac{2u}{e^x}:du = frac{2u}{u^2 - 1}:du
        end{equation}



        Thus:



        begin{align}
        I &= int e^x sqrt{e^x + 1}:dx = int left(u^2 - 1right)sqrt{u^2}frac{2u}{u^2 - 1}:du \
        &= int 2u^2 :du = frac{u^3}{3} + C = frac{1}{3}left(e^x + 1 right)^frac{3}{2} + C
        end{align}



        Where $C$ is the constant of integration.






        share|cite|improve this answer









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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          $(1)$ Because when you make the substitution $color{purple}{u = e^x+1}$, taking the derivative of $u$ yields $frac{du}{dx} = e^x$. Hence, $color{blue}{e^xdx = frac{du}{dx}dx = du}$. The integral can therefore be rewritten:



          $$int color{blue}{e^x}sqrt{color{purple}{e^x+1}} color{blue}{dx} implies intsqrt{color{purple}{u}}color{blue}{du}$$



          $(2)$ In the previous part, note that we get $frac{du}{dx} = e^x$. This can be rewritten as $frac{du}{e^x} = dx$, and $e^x$ in the denominator can be written as $e^{-x}$ in the numerator, so $dx = e^{-x}du$.



          $(3)$ It’s certainly not necessary. It makes the integral easier to evaluate.






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            $(1)$ Because when you make the substitution $color{purple}{u = e^x+1}$, taking the derivative of $u$ yields $frac{du}{dx} = e^x$. Hence, $color{blue}{e^xdx = frac{du}{dx}dx = du}$. The integral can therefore be rewritten:



            $$int color{blue}{e^x}sqrt{color{purple}{e^x+1}} color{blue}{dx} implies intsqrt{color{purple}{u}}color{blue}{du}$$



            $(2)$ In the previous part, note that we get $frac{du}{dx} = e^x$. This can be rewritten as $frac{du}{e^x} = dx$, and $e^x$ in the denominator can be written as $e^{-x}$ in the numerator, so $dx = e^{-x}du$.



            $(3)$ It’s certainly not necessary. It makes the integral easier to evaluate.






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              $(1)$ Because when you make the substitution $color{purple}{u = e^x+1}$, taking the derivative of $u$ yields $frac{du}{dx} = e^x$. Hence, $color{blue}{e^xdx = frac{du}{dx}dx = du}$. The integral can therefore be rewritten:



              $$int color{blue}{e^x}sqrt{color{purple}{e^x+1}} color{blue}{dx} implies intsqrt{color{purple}{u}}color{blue}{du}$$



              $(2)$ In the previous part, note that we get $frac{du}{dx} = e^x$. This can be rewritten as $frac{du}{e^x} = dx$, and $e^x$ in the denominator can be written as $e^{-x}$ in the numerator, so $dx = e^{-x}du$.



              $(3)$ It’s certainly not necessary. It makes the integral easier to evaluate.






              share|cite|improve this answer









              $endgroup$



              $(1)$ Because when you make the substitution $color{purple}{u = e^x+1}$, taking the derivative of $u$ yields $frac{du}{dx} = e^x$. Hence, $color{blue}{e^xdx = frac{du}{dx}dx = du}$. The integral can therefore be rewritten:



              $$int color{blue}{e^x}sqrt{color{purple}{e^x+1}} color{blue}{dx} implies intsqrt{color{purple}{u}}color{blue}{du}$$



              $(2)$ In the previous part, note that we get $frac{du}{dx} = e^x$. This can be rewritten as $frac{du}{e^x} = dx$, and $e^x$ in the denominator can be written as $e^{-x}$ in the numerator, so $dx = e^{-x}du$.



              $(3)$ It’s certainly not necessary. It makes the integral easier to evaluate.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 10 at 20:24









              KM101KM101

              6,0861525




              6,0861525























                  0












                  $begingroup$

                  If $e^{x}-1 =u rightarrow e^xdx=du$ (I suggest writing it this way) so the integral after the substitution looks like this:
                  $int (e^x sqrt{e^x-1})dx=int(sqrt{u})du=cfrac{2u^{3/2}}{3} + C=cfrac{2(e^x-1)^{3/2}}{3}+C$






                  share|cite|improve this answer











                  $endgroup$


















                    0












                    $begingroup$

                    If $e^{x}-1 =u rightarrow e^xdx=du$ (I suggest writing it this way) so the integral after the substitution looks like this:
                    $int (e^x sqrt{e^x-1})dx=int(sqrt{u})du=cfrac{2u^{3/2}}{3} + C=cfrac{2(e^x-1)^{3/2}}{3}+C$






                    share|cite|improve this answer











                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      If $e^{x}-1 =u rightarrow e^xdx=du$ (I suggest writing it this way) so the integral after the substitution looks like this:
                      $int (e^x sqrt{e^x-1})dx=int(sqrt{u})du=cfrac{2u^{3/2}}{3} + C=cfrac{2(e^x-1)^{3/2}}{3}+C$






                      share|cite|improve this answer











                      $endgroup$



                      If $e^{x}-1 =u rightarrow e^xdx=du$ (I suggest writing it this way) so the integral after the substitution looks like this:
                      $int (e^x sqrt{e^x-1})dx=int(sqrt{u})du=cfrac{2u^{3/2}}{3} + C=cfrac{2(e^x-1)^{3/2}}{3}+C$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 10 at 20:29

























                      answered Jan 10 at 20:21









                      Rafał SzypulskiRafał Szypulski

                      35015




                      35015























                          0












                          $begingroup$

                          The way I learnt it more then sixty years ago.



                          $$I=int e^x sqrt{e^x+1},dx$$
                          $$e^x+1=uimplies x=log(u-1) implies dx=frac{du}{u-1}$$ making
                          $$I=int (u-1), sqrt u ,frac{du}{u-1}=int sqrt u ,du=cdots$$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            The way I learnt it more then sixty years ago.



                            $$I=int e^x sqrt{e^x+1},dx$$
                            $$e^x+1=uimplies x=log(u-1) implies dx=frac{du}{u-1}$$ making
                            $$I=int (u-1), sqrt u ,frac{du}{u-1}=int sqrt u ,du=cdots$$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              The way I learnt it more then sixty years ago.



                              $$I=int e^x sqrt{e^x+1},dx$$
                              $$e^x+1=uimplies x=log(u-1) implies dx=frac{du}{u-1}$$ making
                              $$I=int (u-1), sqrt u ,frac{du}{u-1}=int sqrt u ,du=cdots$$






                              share|cite|improve this answer









                              $endgroup$



                              The way I learnt it more then sixty years ago.



                              $$I=int e^x sqrt{e^x+1},dx$$
                              $$e^x+1=uimplies x=log(u-1) implies dx=frac{du}{u-1}$$ making
                              $$I=int (u-1), sqrt u ,frac{du}{u-1}=int sqrt u ,du=cdots$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 11 at 7:03









                              Claude LeiboviciClaude Leibovici

                              125k1158135




                              125k1158135























                                  0












                                  $begingroup$

                                  Another approach:



                                  begin{equation}
                                  I = int e^x sqrt{e^x + 1}:dx
                                  end{equation}



                                  Here let $u^2 = e^x + 1$. Employing Implicit Differentiation we see



                                  begin{equation}
                                  frac{d}{dx}left[u^2right] = frac{d}{dx}left[e^x + 1 right] rightarrow 2u frac{du}{dx} = e^x rightarrow dx = frac{2u}{e^x}:du = frac{2u}{u^2 - 1}:du
                                  end{equation}



                                  Thus:



                                  begin{align}
                                  I &= int e^x sqrt{e^x + 1}:dx = int left(u^2 - 1right)sqrt{u^2}frac{2u}{u^2 - 1}:du \
                                  &= int 2u^2 :du = frac{u^3}{3} + C = frac{1}{3}left(e^x + 1 right)^frac{3}{2} + C
                                  end{align}



                                  Where $C$ is the constant of integration.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Another approach:



                                    begin{equation}
                                    I = int e^x sqrt{e^x + 1}:dx
                                    end{equation}



                                    Here let $u^2 = e^x + 1$. Employing Implicit Differentiation we see



                                    begin{equation}
                                    frac{d}{dx}left[u^2right] = frac{d}{dx}left[e^x + 1 right] rightarrow 2u frac{du}{dx} = e^x rightarrow dx = frac{2u}{e^x}:du = frac{2u}{u^2 - 1}:du
                                    end{equation}



                                    Thus:



                                    begin{align}
                                    I &= int e^x sqrt{e^x + 1}:dx = int left(u^2 - 1right)sqrt{u^2}frac{2u}{u^2 - 1}:du \
                                    &= int 2u^2 :du = frac{u^3}{3} + C = frac{1}{3}left(e^x + 1 right)^frac{3}{2} + C
                                    end{align}



                                    Where $C$ is the constant of integration.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Another approach:



                                      begin{equation}
                                      I = int e^x sqrt{e^x + 1}:dx
                                      end{equation}



                                      Here let $u^2 = e^x + 1$. Employing Implicit Differentiation we see



                                      begin{equation}
                                      frac{d}{dx}left[u^2right] = frac{d}{dx}left[e^x + 1 right] rightarrow 2u frac{du}{dx} = e^x rightarrow dx = frac{2u}{e^x}:du = frac{2u}{u^2 - 1}:du
                                      end{equation}



                                      Thus:



                                      begin{align}
                                      I &= int e^x sqrt{e^x + 1}:dx = int left(u^2 - 1right)sqrt{u^2}frac{2u}{u^2 - 1}:du \
                                      &= int 2u^2 :du = frac{u^3}{3} + C = frac{1}{3}left(e^x + 1 right)^frac{3}{2} + C
                                      end{align}



                                      Where $C$ is the constant of integration.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Another approach:



                                      begin{equation}
                                      I = int e^x sqrt{e^x + 1}:dx
                                      end{equation}



                                      Here let $u^2 = e^x + 1$. Employing Implicit Differentiation we see



                                      begin{equation}
                                      frac{d}{dx}left[u^2right] = frac{d}{dx}left[e^x + 1 right] rightarrow 2u frac{du}{dx} = e^x rightarrow dx = frac{2u}{e^x}:du = frac{2u}{u^2 - 1}:du
                                      end{equation}



                                      Thus:



                                      begin{align}
                                      I &= int e^x sqrt{e^x + 1}:dx = int left(u^2 - 1right)sqrt{u^2}frac{2u}{u^2 - 1}:du \
                                      &= int 2u^2 :du = frac{u^3}{3} + C = frac{1}{3}left(e^x + 1 right)^frac{3}{2} + C
                                      end{align}



                                      Where $C$ is the constant of integration.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 13 at 12:49







                                      user150203





































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                                          Uppslagsordet ”Dürer” leder hit. För andra betydelser, se Dürer (olika betydelser). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedr
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