Trick for Solving Double Integral with e^x/(x+1) Function

[thenewbosco]

Perhaps someone can explain how to do, or a trick to solve this integral:

\int_{0}^{3}\int_{\sqrt{y+1}}^{2} \frac{e^x}{x+1} dx dy

i don't know if \frac{e^x}{x+1} has an elementary antideriv. so there should be some trick to solving this i guess...

 

[Hurkyl]

You have to use '_' and '^' to set the limits on an integral. e.g.

\int_{lower}^{upper}

And you forgot to put your `dx's and `dy's in the problem.

The first thing I think of when I'm stumped on a double integral is to reverse the order of integration to see if that helps. Have you tried that?

 

[thenewbosco]

right there it is correctly formatted. but by reversing the order, you will have a y in the final answer, not like it helps to reverse in this case anyways

 

[Hurkyl]

Well, when you reverse the order, the limits change, right? You won't know if it helps until you try it!

 

[buzzmath]

use integration by parts.

 

[thenewbosco]

what do you choose for u and dv? i cannot simplify it?

 

[buzzmath]

try using u = x+l and dv=e^xdx then du=dx and v=e^x
do you know the formula for integration by parts?

 

[shmoe]

Well, when you reverse the order, the limits change, right? You won't know if it helps until you try it!

Have you, thenewbosco, followed this advice yet?

 

[thenewbosco]

if you try this you get 3*e^x/(x+1) which doesn't really help any?

 

[shmoe]

When you change the order of integration, you have to be careful with the limits. It might help if you sketch the region you are integrating over. What is the integral after you change the order?

 

[HallsofIvy]

try using u = x+l and dv=e^xdx then du=dx and v=e^x
do you know the formula for integration by parts?

Since the problem is
\int \frac{e^x dx}{x+1}
rather than
\int (x+1)e^x dx
that doesn't help at all.


Changing the order of integration is the way to go.

Thenewbosco, have you sketched the area over which you are integrating? That's the best way to find the new limits of integration. Once you've done that you can use buzzmath's suggestion of integration by parts but with u= x-1, not x+1!