## U sub of a real double integral

1. The problem statement, all variables and given/known data

∫∫ 1 / (2x + 3y), R = [0,1] x [1,2]

2. Relevant equations

- Iterated integrals
- u sub

3. The attempt at a solution

Here is my attempt at solving this (I must be screwing up on the algebra)

Integrating with respect to x
u = 2x, dy = 2

u^-2 du

u^-2 = - (1/u) [double chcked with wolfram alpha]
∫∫ (2x + 3y) ^ -2 -> -2 ∫ -(1/u) (2x + 3y) ^ -1

This is the part that stumps me....Sorry if I'm being vague. This is from Paul Online Notes Calculus III (http://tutorial.math.lamar.edu/Class..._Iterated_Ex1d)

THe part I want to understand is how they got -1/2 * ...... (2x+3y) ^ -1

Thanks!
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 Quote by selig5560 1. The problem statement, all variables and given/known data ∫∫ 1 / (2x + 3y), R = [0,1] x [1,2]
For starters, you didn't copy the problem correctly. Should be$$\int_1^2\int_0^1 \frac 1 {(2x+3y)^2}\, dxdy$$
 2. Relevant equations - Iterated integrals - u sub 3. The attempt at a solution Here is my attempt at solving this (I must be screwing up on the algebra) Integrating with respect to x u = 2x, dy = 2 u^-2 du u^-2 = - (1/u) [double chcked with wolfram alpha] ∫∫ (2x + 3y) ^ -2 -> -2 ∫ -(1/u) (2x + 3y) ^ -1 This is the part that stumps me....Sorry if I'm being vague. This is from Paul Online Notes Calculus III (http://tutorial.math.lamar.edu/Class..._Iterated_Ex1d) THe part I want to understand is how they got -1/2 * ...... (2x+3y) ^ -1 Thanks!
Let ##u=2x+3y## and ##du=2dx##. y is constant for the inner integral. The inner integral becomes$$\int u^{-2} \frac 1 2 du = \frac 1 2 \frac {u^{-1}}{-1}$$