Double integral involving (x+y)

[medicalphysicsguy]

I actually think my problem is not understanding how to handle the x+y in parentheses here.

\(\displaystyle \int_1^2\int_0^2\frac{dydx}{(x+y)^2}\)

I just treat x as a constant so I don't see it having any impact on the first integral:

\(\displaystyle \int_1^2-(x+y)^{-1}\rbrack_0^2dx\)

leading to:

\(\displaystyle \int_1^2(-(x+2)^{-1}+x^{-1})dx\)

or:

\(\displaystyle -ln(x+2)+ln(x)\rbrack_1^2\)

\(\displaystyle -ln(4)+ln(2)-ln(3)+ln(1)\)

Unless I have forgotten some nifty identity, this does not get me the answer of \(\displaystyle ln\frac{3}{2}\).Have I not handled the parentheses right?

Thanks, mpg

 

[srmichael]

I actually think my problem is not understanding how to handle the x+y in parentheses here.

\(\displaystyle \int_1^2\int_0^2\frac{dydx}{(x+y)^2}\)

I just treat x as a constant so I don't see it having any impact on the first integral:

\(\displaystyle \int_1^2-(x+y)^{-1}\rbrack_0^2dx\)

leading to:

\(\displaystyle \int_1^2(-(x+2)^{-1}+x^{-1})dx\)

or:

\(\displaystyle -ln(x+2)+ln(x)\rbrack_1^2\)

\(\displaystyle -ln(4)+ln(2)-ln(3)+ln(1)\)

Unless I have forgotten some nifty identity, this does not get me the answer of \(\displaystyle ln\frac{3}{2}\).Have I not handled the parentheses right?

Thanks, mpg

Close:

\(\displaystyle -ln(x+2)+ln(x)\rbrack_1^2\)

\(\displaystyle -ln(4)+ln(2)-[-ln(3)+ln(1)]\) (You forgot to subtract when you plugged in 1)

\(\displaystyle -ln(4)+ln(2)+ln(3)-ln(1)\)

\(\displaystyle -ln(2^2)+ln(2)+ln(3)-0\)

\(\displaystyle -2ln(2)+ln(2)+ln(3)\)

\(\displaystyle ln(3)-ln(2)\)

\(\displaystyle ln(\frac{3}{2})\)