Proof of exponential function integral











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I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule



$$int a^x dx= frac{a^x}{ln(a)}+C$$



would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.






calculus integration proof-writing exponential-function







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  • 5




    It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    7 hours ago










  • @Jakobian : The only real answer. +1.
    – MPW
    7 hours ago















up vote
2
down vote

favorite
1












I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule



$$int a^x dx= frac{a^x}{ln(a)}+C$$



would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.






calculus integration proof-writing exponential-function







share|cite|improve this question




















  • 5




    It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    7 hours ago










  • @Jakobian : The only real answer. +1.
    – MPW
    7 hours ago













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule



$$int a^x dx= frac{a^x}{ln(a)}+C$$



would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.






calculus integration proof-writing exponential-function







share|cite|improve this question















I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule



$$int a^x dx= frac{a^x}{ln(a)}+C$$



would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.






calculus integration proof-writing exponential-function




calculus integration proof-writing exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago

























asked 7 hours ago









Math Bob

138




138








  • 5




    It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    7 hours ago










  • @Jakobian : The only real answer. +1.
    – MPW
    7 hours ago














  • 5




    It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    7 hours ago










  • @Jakobian : The only real answer. +1.
    – MPW
    7 hours ago








5




5




It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
– Jakobian
7 hours ago




It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
– Jakobian
7 hours ago












@Jakobian : The only real answer. +1.
– MPW
7 hours ago




@Jakobian : The only real answer. +1.
– MPW
7 hours ago










3 Answers
3






active

oldest

votes

















up vote
4
down vote



accepted










I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



$$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



If so, then note:



$$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



This is mostly just manipulation of various logarithm properties: namely,



$$e^{ln(x)} = x$$
$$ln(a^b) = b ln(a)$$



Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.






share|cite|improve this answer





















  • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
    – Math Bob
    7 hours ago












  • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
    – Eevee Trainer
    6 hours ago










  • Okay. I have taken this in so thanks for the answer!
    – Math Bob
    3 hours ago


















up vote
1
down vote













Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



$$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



$$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



$$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




$$int a^xdx=frac{a^x}{ln(a)}+k$$







share|cite|improve this answer




























    up vote
    1
    down vote













    Just differentiate the right hand side and see what you get. Note that
    $$
    frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
    $$
    where we used the chain rule in the second equality.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



      $$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



      If so, then note:



      $$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



      This is mostly just manipulation of various logarithm properties: namely,



      $$e^{ln(x)} = x$$
      $$ln(a^b) = b ln(a)$$



      Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.






      share|cite|improve this answer





















      • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
        – Math Bob
        7 hours ago












      • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
        – Eevee Trainer
        6 hours ago










      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        3 hours ago















      up vote
      4
      down vote



      accepted










      I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



      $$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



      If so, then note:



      $$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



      This is mostly just manipulation of various logarithm properties: namely,



      $$e^{ln(x)} = x$$
      $$ln(a^b) = b ln(a)$$



      Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.






      share|cite|improve this answer





















      • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
        – Math Bob
        7 hours ago












      • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
        – Eevee Trainer
        6 hours ago










      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        3 hours ago













      up vote
      4
      down vote



      accepted







      up vote
      4
      down vote



      accepted






      I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



      $$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



      If so, then note:



      $$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



      This is mostly just manipulation of various logarithm properties: namely,



      $$e^{ln(x)} = x$$
      $$ln(a^b) = b ln(a)$$



      Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.






      share|cite|improve this answer












      I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



      $$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



      If so, then note:



      $$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



      This is mostly just manipulation of various logarithm properties: namely,



      $$e^{ln(x)} = x$$
      $$ln(a^b) = b ln(a)$$



      Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 7 hours ago









      Eevee Trainer

      3,340225




      3,340225












      • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
        – Math Bob
        7 hours ago












      • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
        – Eevee Trainer
        6 hours ago










      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        3 hours ago


















      • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
        – Math Bob
        7 hours ago












      • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
        – Eevee Trainer
        6 hours ago










      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        3 hours ago
















      OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
      – Math Bob
      7 hours ago






      OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
      – Math Bob
      7 hours ago














      Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
      – Eevee Trainer
      6 hours ago




      Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
      – Eevee Trainer
      6 hours ago












      Okay. I have taken this in so thanks for the answer!
      – Math Bob
      3 hours ago




      Okay. I have taken this in so thanks for the answer!
      – Math Bob
      3 hours ago










      up vote
      1
      down vote













      Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



      $$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



      Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



      $$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



      From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



      $$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




      $$int a^xdx=frac{a^x}{ln(a)}+k$$







      share|cite|improve this answer

























        up vote
        1
        down vote













        Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



        $$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



        Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



        $$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



        From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



        $$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




        $$int a^xdx=frac{a^x}{ln(a)}+k$$







        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



          $$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



          Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



          $$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



          From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



          $$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




          $$int a^xdx=frac{a^x}{ln(a)}+k$$







          share|cite|improve this answer












          Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



          $$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



          Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



          $$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



          From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



          $$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




          $$int a^xdx=frac{a^x}{ln(a)}+k$$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          mrtaurho

          2,9901928




          2,9901928






















              up vote
              1
              down vote













              Just differentiate the right hand side and see what you get. Note that
              $$
              frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
              $$
              where we used the chain rule in the second equality.






              share|cite|improve this answer

























                up vote
                1
                down vote













                Just differentiate the right hand side and see what you get. Note that
                $$
                frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
                $$
                where we used the chain rule in the second equality.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Just differentiate the right hand side and see what you get. Note that
                  $$
                  frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
                  $$
                  where we used the chain rule in the second equality.






                  share|cite|improve this answer












                  Just differentiate the right hand side and see what you get. Note that
                  $$
                  frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
                  $$
                  where we used the chain rule in the second equality.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  Foobaz John

                  20.4k41250




                  20.4k41250






























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                      Bundesstraße 106

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                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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                      Verónica Boquete

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                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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