`int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx.` Hence evaluate : `int_(a)^(b)(f(x))/(f(x)+f(a+b-x))dx.`

Explanation

`intf(x)dx=int_(a)^(b)f(a+b-x)Dx`
Since we know that the following properties,
`int_(a)^(b)f(x)dx=-int_(b)^(a)f(x)dx" …(i)"`
and `" "int_(a)^(b)f(x)=int_(a)^(b)f(t)dt" …(ii)"`
Consider, `int_(a)^(b)f(x)dx`
Putting `x=a+b - t." "(because t = a+b-x)`
`rArr" "dx=-dt`
When `x=a,` then `t=a+b-a=b`
When `x=b`, then `t=a+b-b=a`
`therefore int_(a)^(b)f(x)dx=int_(b)^(a)f(a+b-t)(-dt)`
`=-int_(b)^(a)f(a+h-t)dt`
`=int_(a)^(b)f(a+b-t)dt" [By equation (i)]"`
`=int_(a)^(b)f(a+b-t)dx" [By equation (ii)]"`
And now,
`int_(a)^(b)(f(x))/(f(x)+f(a+b-x))dx`
`"Let I"=int_(a)^(b)(f(x))/(f(x)+f(a+b-x))dx" ...(iii)"`
And`" I"=int_(a)^(b)(f(a+b-x))/(f(a+b-x)+f(x))" ...(iv)"`
Adding equations (i) and (ii), we get
`2I=int_(a)^(b)(f(x)+f(a+b-x))/(f(x)+f(a+b-x))dx`
`=int_(a)^(b)1.dx`
`=[x]_(a)^(b)`
`therefore" "2I=b-a`
`therefore" "I=(b-a)/(2)`
`therefore" "=int_(a)^(b)(f(x))/(f(x)+f(A+b-x))dx`
`=(b-a)/(2)`