U sub of a real double integral

[selig5560]

Homework Statement

∫∫ 1 / (2x + 3y), R = [0,1] x [1,2]

Homework Equations

- Iterated integrals
- u sub

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/Classes/CalcIII/IteratedIntegrals.aspx#MultInt_Iterated_Ex1d)

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

Thanks!

 

[LCKurtz]

Homework Statement

∫∫ 1 / (2x + 3y), R = [0,1] x [1,2]

For starters, you didn't copy the problem correctly. Should be$$
\int_1^2\int_0^1 \frac 1 {(2x+3y)^2}\, dxdy$$

Homework Equations

- Iterated integrals
- u sub

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/Classes/CalcIII/IteratedIntegrals.aspx#MultInt_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$$
\int u^{-2} \frac 1 2 du = \frac 1 2 \frac {u^{-1}}{-1}$$