`Iff(x)=x+int_0^1t(x+t)f(t)dt ,` `t h e nfin dt h ev a l u eoft h ed efin i t ein t egr a lint_0^1f(x)dxdot`

Explanation

`f(x)=x+x int_(0)^(1)tf(t)dt+int_(0)^(1)t^(2)f(t)dt`
`:.f(x)=x(1+A)+B`
where `A=int_(0)^(1)tf(t)dt` and `B=int_(0)^(1)t^(2)f(t)dt`
`A=int_(0)^(1)t[t(1+A)+B]dt=[(1+A)(t^(3))/3+B(t^(2))/2]_(0)^(1)`
`:.A+(1+A)/3+B/2`
or `4A-3B=2` ..........1
`B=int_(0)^(1)t^(2)[t(1+A)+B]dt`
`=[(t^(4)(1+A))/4+(Br^(3))/3]_(0)^(1)`
`=(1+A)/4+B/3` ltbr or `8B-3A=3`..............2
Solving 1 and 2 we get `A=25/23` and `B=18/23`
`:.f(x)=(48x+18)/23`