# Solve Antiderivative of xe^x | Step-by-Step Guide

• dfklajsdfald
In summary: That's not how it works -- you are applying the power rule for a function for which it doesn't apply.
dfklajsdfald
Member warned that an effort must be shown

## Homework Statement

find the anti-derivative of xe ^x so its x to the power of e to the power of x

## The Attempt at a Solution

i have 0 idea where to even start. this was a question on my quiz today

Hi dfk,

No homework. Still an attempt on your part required $$F'(x) = x^{e^x} \ \Rightarrow \ F(x) = ??$$

well i know that the derivative of ex is ex so the anti derivative of ex is ex and then my attempt would be exx but i don't know if that's right or even a good start

BvU said:
Hi dfk,

No homework. Still an attempt on your part required $$F'(x) = x^{e^x} \ \Rightarrow \ F(x) = ??$$
actually scratch the last part. my attempt would be xe^x +1 becacuse normally for any function x u go x and add one to the power and divide by it

dfklajsdfald said:
actually scratch the last part. my attempt would be xe^x +1 becacuse normally for any function x u go x and add one to the power and divide by it
So is ##\frac d{dx}(x^{e^x + 1}) = x^{e^x}##?

For a power function, such as ##x^n##, the derivative is ##nx^{n - 1}##, but is that what you have here?

Do you confirm that the formula in #2 is the one you meant?

Then you don't need to ask us whether a proposed solution is correct, surely you can differentiate?

If then you find it wrong, It may at least suggest something else to try. Or maybe some other direction to look.

epenguin said:
Do you confirm that the formula in #2 is the one you meant?

Then you don't need to ask us whether a proposed solution is correct, surely you can differentiate?

If then you find it wrong, It may at least suggest something else to try. Or maybe some other direction to look.

i don't know what ur asking, the question is find the antiderivative of the function x to the power of e to the power of x as stated at the start. I am not asking whether or not a solution is correct I am asking how abouts would i go to try and get the antiderivative because I've tried multiple ways (one which i have stated in a different reply) and can't seem to get there. I am only asking for guidance not for an answer of whether or not my solution is correct.

Mark44 said:
So is ##\frac d{dx}(x^{e^x + 1}) = x^{e^x}##?

For a power function, such as ##x^n##, the derivative is ##nx^{n - 1}##, but is that what you have here?
what i meant was that it was my attempt at finding the antiderivative of xex. i know that the derivative of ex is ex therefor the anti derivative would be the same. so then for derivatives with base x you go xn-1 times the derivative of the power but since i need the anti derivative wouldn't i do xn+1 divided by the power instead? so wouldn't it be xe^x +1 all divided by ex(the plus one is added to ex not to the entire function I am just having trouble showing it up on here properly)

dfklajsdfald said:
what i meant was that it was my attempt at finding the antiderivative of xex. i know that the derivative of ex is ex therefor the anti derivative would be the same. so then for derivatives with base x you go xn-1 times the derivative of the power but since i need the anti derivative wouldn't i do xn+1 divided by the power instead?
That's not how it works -- you are applying the power rule for a function for which it doesn't apply.
Power rule: ##\frac d{dx} x^n = nx^{n - 1}##
Here the exponent, n, is a constant. For your function the exponent, ##e^x + 1## is not a constant.

By your reasoning, ##\frac d{dx} x^x =xx^{x - 1} = x^x##, which isn't true. The derivative of ##x^x## is actually ##x^x(\ln x + 1)##.
dfklajsdfald said:
so wouldn't it be xe^x +1 all divided by ex(the plus one is added to ex not to the entire function I am just having trouble showing it up on here properly)

cnh1995
dfklajsdfald said:
what i meant was that it was my attempt at finding the antiderivative of xex. i know that the derivative of ex is ex therefor the anti derivative would be the same. so then for derivatives with base x you go xn-1 times the derivative of the power but since i need the anti derivative wouldn't i do xn+1 divided by the power instead? so wouldn't it be xe^x +1 all divided by ex(the plus one is added to ex not to the entire function I am just having trouble showing it up on here properly)

To get the derivative, write ##x^{e^x}= e^{f(x)}##, where ##f(x) = \ln(x) e^x##, and use the chain rule.

dfklajsdfald said:
i don't know what ur asking, the question is find the antiderivative of the function x to the power of e to the power of x as stated at the start. I am not asking whether or not a solution is correct I am asking how abouts would i go to try and get the antiderivative because I've tried multiple ways (one which i have stated in a different reply) and can't seem to get there. I am only asking for guidance not for an answer of whether or not my solution is correct.

I thought and rereading #3 it still looks to me you were asking whether an answer was correct. We do often get to ask that question for integrals and have to point out that checking is the easy part - and also necessary. It seems to me you have now got some guidance to work on.

Ray Vickson said:
To get the derivative, write ##x^{e^x}= e^{f(x)}##, where ##f(x) = \ln(x) e^x##, and use the chain rule.
ya but the question is asking for antiderivative

dfklajsdfald said:
ya but the question is asking for antiderivative
You claimed earlier that the antiderivative was ##x^{e^x + 1}##. If this is correct, you should be able to differentiate this function to get your original integrand. That's what we're helping you with.

On the other hand, if you differentiate ##x^{e^x + 1}##, and DON'T get ##x^{e^x}##, then that tells you that your purported antiderivative is incorrect.

Mark44 said:
You claimed earlier that the antiderivative was ##x^{e^x + 1}##. If this is correct, you should be able to differentiate this function to get your original integrand. That's what we're helping you with.

On the other hand, if you differentiate ##x^{e^x + 1}##, and DON'T get ##x^{e^x}##, then that tells you that your purported antiderivative is incorrect.
ohh okay that makes more sense. and i wasnt claiming that that's what the antiderivative was. i was saying that was how i was trying to solve the antiderviative but was getting the wrong answer so i needed guidance to go abouts finding the antiderivative correctly. you asked what my initial attempt was which is why i responded with that, i know its wrong but u wanted to know what i did so that's why i showed that

dfklajsdfald said:
ohh okay that makes more sense. and i wasnt claiming that that's what the antiderivative was.
In post #4 you said:
dfklajsdfald said:
actually scratch the last part. my attempt would be xe^x +1 becacuse normally for any function x u go x and add one to the power and divide by it
It looks to me like you were claiming that ##x^{e^x + 1}## is the antiderivative of ##x^{e^x}##.

dfklajsdfald said:
i was saying that was how i was trying to solve the antiderviative but was getting the wrong answer so i needed guidance to go abouts finding the antiderivative correctly. you asked what my initial attempt was which is why i responded with that, i know its wrong but u wanted to know what i did so that's why i showed that

Please don't use "text speak" here at PF. In the rules for this site (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/ ), under General Posting Guidelines, it says"
SMS messaging shorthand ("text-message-speak"), such as using "u" for "you", "please" for "please", or "wanna" for "want to" is not acceptable.

Mark44 said:
In post #4 you said:

It looks to me like you were claiming that ##x^{e^x + 1}## is the antiderivative of ##x^{e^x}##.
Please don't use "text speak" here at PF. In the rules for this site (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/ ), under General Posting Guidelines, it says"
i said my "attempt" i never claimed anything?? but anyway nevermind i'll figure it out on my own i just wanted some help and guidance not an argument or a forum of whose wrong whose right i'll delete this shortly

## 1. How do I solve the antiderivative of xe^x?

To solve the antiderivative of xe^x, you can use integration by parts. First, set u = x and dv = e^x. This will give you du = dx and v = e^x. Then, use the formula ∫udv = uv - ∫vdu to solve for the antiderivative. In this case, the antiderivative would be xe^x - e^x + C.

## 2. Can I solve the antiderivative of xe^x without using integration by parts?

Yes, there are alternative methods to solve the antiderivative of xe^x. One method is to use the substitution technique. Set u = x and du = dx, then substitute these into the integral to get ∫xe^x dx = ∫ue^u du. This can then be solved using the power rule for integration.

## 3. What if there is a constant in front of the x in the antiderivative, such as 3xe^x?

If there is a constant in front of the x, such as 3xe^x, you can still use integration by parts. Set u = 3x and dv = e^x, then follow the same steps as before to solve for the antiderivative. The result would be 3xe^x - 3e^x + C.

## 4. Is there a general formula for solving antiderivatives of exponential functions?

Yes, there is a general formula for solving antiderivatives of exponential functions. For the function f(x) = ae^(bx), the antiderivative is given by (1/b)ae^(bx) + C. This formula can be applied to solve for antiderivatives of various exponential functions such as xe^x, 3e^x, and so on.

## 5. Are there any special cases when solving the antiderivative of xe^x?

Yes, there is a special case when solving the antiderivative of xe^x. If the function is f(x) = e^x, then the antiderivative is simply e^x + C. This is because the derivative of e^x is also e^x, so the two functions cancel each other out.

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