Proof of exponential function integral

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2

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.

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  • 5

    It’s quite simple. $(a^x)’ = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    Dec 16 at 23:21

  • @Jakobian : The only real answer. +1.
    – MPW
    Dec 16 at 23:46

2

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.

share|cite|improve this question

  • 5

    It’s quite simple. $(a^x)’ = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    Dec 16 at 23:21

  • @Jakobian : The only real answer. +1.
    – MPW
    Dec 16 at 23:46

2

2

2

0

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.

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

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edited Dec 16 at 23:54

asked Dec 16 at 23:16

Math Bob

138

138

  • 5

    It’s quite simple. $(a^x)’ = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    Dec 16 at 23:21

  • @Jakobian : The only real answer. +1.
    – MPW
    Dec 16 at 23:46

  • 5

    It’s quite simple. $(a^x)’ = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    Dec 16 at 23:21

  • @Jakobian : The only real answer. +1.
    – MPW
    Dec 16 at 23:46

5

5

It’s quite simple. $(a^x)’ = a^xcdot ln(a)$, and the rule follows.
– Jakobian
Dec 16 at 23:21

It’s quite simple. $(a^x)’ = a^xcdot ln(a)$, and the rule follows.
– Jakobian
Dec 16 at 23:21

@Jakobian : The only real answer. +1.
– MPW
Dec 16 at 23:46

@Jakobian : The only real answer. +1.
– MPW
Dec 16 at 23:46

3 Answers
3

active

oldest

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4

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
    Dec 16 at 23:51

  • 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
    Dec 17 at 0:18

  • Okay. I have taken this in so thanks for the answer!
    – Math Bob
    Dec 17 at 4:04

1

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

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

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

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    1

    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
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      4

      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
        Dec 16 at 23:51

      • 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
        Dec 17 at 0:18

      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        Dec 17 at 4:04

      4

      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
        Dec 16 at 23:51

      • 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
        Dec 17 at 0:18

      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        Dec 17 at 4:04

      4

      4

      4

      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 Dec 16 at 23:23

      Eevee Trainer

      4,048529

      4,048529

      • 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
        Dec 16 at 23:51

      • 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
        Dec 17 at 0:18

      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        Dec 17 at 4:04

      • 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
        Dec 16 at 23:51

      • 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
        Dec 17 at 0:18

      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        Dec 17 at 4:04

      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
      Dec 16 at 23:51

      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
      Dec 16 at 23:51

      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
      Dec 17 at 0:18

      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
      Dec 17 at 0:18

      Okay. I have taken this in so thanks for the answer!
      – Math Bob
      Dec 17 at 4:04

      Okay. I have taken this in so thanks for the answer!
      – Math Bob
      Dec 17 at 4:04

      1

      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

        1

        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

          1

          1

          1

          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 Dec 16 at 23:25

          mrtaurho

          3,3962932

          3,3962932

              1

              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

                1

                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

                  1

                  1

                  1

                  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 Dec 16 at 23:44

                  Foobaz John

                  20.9k41250

                  20.9k41250

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