# Proof of exponential function integral

Clash Royale CLAN TAG#URR8PPP

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.

• 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

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.

• 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

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.

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

edited Dec 16 at 23:54

Math Bob

138

138

• 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

• 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

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

active

oldest

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.

• 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

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

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

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

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.

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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.

• 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

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.

• 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

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.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

mrtaurho

3,3962932

3,3962932

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.

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.

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.

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.

Foobaz John

20.9k41250

20.9k41250

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