These were posted on another forum.

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- February 25th 2011, 12:09 AMRandom Variableindefinite integrals (lots of them)
These were posted on another forum.

1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

12) - February 25th 2011, 12:30 AMProve It
- February 25th 2011, 12:50 AMTheCoffeeMachine
- February 25th 2011, 12:58 AMProve It
.

Make the substitution and the integral becomes

. - February 25th 2011, 07:21 AMRandom VariableQuote:

Surely an MHF Expert knows to only post 2 questions per thread...

- February 25th 2011, 07:58 AMProve It

Let and the integral becomes

.

Now let and the integral becomes

.

This can probably be simplified more using logarithmic equivalents, but I think this is sufficient... - February 25th 2011, 08:04 AMRandom Variable
3)

let

so

- February 25th 2011, 09:36 AMTheEmptySet
4)

gives

- February 25th 2011, 10:50 AMTheEmptySet
6)

let

This gives

- February 25th 2011, 11:05 AMRandom Variable

let

let

- February 25th 2011, 11:11 AMTheEmptySet
5)

Integrate by parts with

and

This gives

- February 25th 2011, 01:56 PMTheCoffeeMachine
An altenative solution for #5:

- February 25th 2011, 08:51 PMTheCoffeeMachine