Is This Integration by Parts Approach Correct for Solving ∫(xe^x + e^x)dx?

In summary, the conversation discusses integration by parts and a substitution method for solving the integral of e^x+e^x. The correct solution is given as e^(e^x) +c. The conversation also suggests breaking the integral into two separate integrals for easier calculation.
  • #1
afcwestwarrior
457
0

Homework Statement


∫e^x+e^x




Homework Equations


∫u dv= uv- ∫v du


The Attempt at a Solution


u= x+e^x
du= e^x

so it would be e^u

integral = e^u

= e^(e^x) +c is that correct, i know the answer is but what i just did
 
Physics news on Phys.org
  • #2
Did you mistype the integral? [tex]\int 2e^x dx[/tex]?
 
  • #3
afcwestwarrior said:

Homework Statement


∫e^x+e^x
? Surely you don't mean [itex]2\int e^x dx[/itex]?




Homework Equations


∫u dv= uv- ∫v du


The Attempt at a Solution


u= x+e^x
du= e^x
Are you now saying the problem is [itex]\int (xe^x+ e^x)dx[/itex]?
Then you are not using integration by parts, you are using a simple substitution. Yes, [itex]\int (xe^x+ e^x)dx= \int (x+ e^x)e^x dx[/itex]. If you let u= x+ ex, then du= (1+ e^x) dx, not just ex dx. And please by sure to include the "dx" in the integral; that may be part of what is confusing you.

so it would be e^u

integral = e^u
= e^(e^x) +c is that correct, i know the answer is but what i just did
Well, you can always check an integration yourself by differentiating.
[tex]\frac{d}{dx}\left(e^{e^x}\right)= \frac{de^u}{du}\frac{de^x}{dx}[/tex]
with u= ex
[tex]= (e^u)(e^x)= (e^{e^x})(e^x)[/tex]
Which is not what you started with.
Did you consider just doing the two integrals separately?
[tex]\int (xe^x+ e^x)dx= \int xe^x dx+ \int e^x dx[/itex]
You should be able to do the second of those directly and the first is a simple integration by parts.
 

FAQ: Is This Integration by Parts Approach Correct for Solving ∫(xe^x + e^x)dx?

What is the general formula for integration by parts?

The general formula for integration by parts is ∫u dv = uv - ∫v du, where u is the first function and dv is the second function. This formula is used to solve integrals where one function is difficult to integrate, but the product of two functions is easier to integrate.

How do you choose which function to use as u and which to use as dv?

The u-substitution rule can be used to determine which function to use as u. This rule states that the function that appears first when performing the derivative (from left to right) should be used as u. The remaining function will then be used as dv.

Can integration by parts be used on all integrals?

No, integration by parts can only be used on integrals where the product of two functions is present. It cannot be used on integrals that do not have this form.

How many times can integration by parts be used on a single integral?

Integration by parts can be used multiple times on a single integral if necessary. Each time it is used, the integral becomes simpler and more manageable.

Are there any specific rules to follow when using integration by parts?

Yes, there are certain rules that should be followed when using integration by parts. These include using the u-substitution rule to determine which function to use as u, choosing the correct dv function, and applying the formula correctly. It is also important to pay attention to any special cases, such as when the integral becomes an infinite series or when it becomes a repeated integral.

Similar threads

Back
Top