In this problem, we must first do method of substitution, and separate X from the integrand, then we can get the derivative.

Let tx=u, t=u/x, dt= 1/x du, t: 0-> 1, u: 0-> x

The original equation is: ∫ [0-> x] f (u) du/x =1/2f (x)+1

Namely: 2 ∫ [0->; x] f(u)du=xf(x)+x

The derivative of x on both sides is 2f(x)=f(x)+xf '(x)+ 1.

F(x)=xf '(x)+ 1。

That is df(x)/(f(x)- 1)=xdx.

ln|f(x)- 1|=( 1/2)x^2+ln|c|

F (x)- 1 = ce (( 1/2) x 2)

Or (f (x)-1) 2 = ce (x 2).