Answers

2014-10-27T18:32:28+05:30
2 f(x)+3 f(1/x)=1/x-2.........................................1
now placing 1/x in place of x
we get
2f(1/x)+3f(x)=x-2.........................2
multiply eq1 by 2 and 2by 3 we get
4f(x)+6f(1/x)=2/x-4.............................3
6f(1/x)+9f(x)=3x-6.....................................4
subtracting botheq
5f(x)=3x-6-2/x+4=(3x^2-2x-2)/x
so..................
f(x)=[3x^2-2x-2]/5x

now
 \int\limits^1_2 {[3x^2-2x-2]}/5x \, dx
=3-2-2-12+4+2
=-7


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