Leibniz Rule for Differentiation of Integrals - Natural Logarithm...?

Suppose ∫ 1/ln(x+u) du = F(u) + C
Then dF(u)/du = 1/ln(x+u)

d/dx { ∫ 1/ln(x+u) du , from u=x to u=x^2 }
= d/dx { F(u) , from u=x to u=x^2 }
= d/dx [ F(x^2) - F(x) ]
= dF(x^2)/dx - dF(x)/dx
= [ dF(x^2)/dx^2 ]*( dx^2/dx ) - 1/ln(x+x)
= [ 1/ln(x+x^2) ]*2x - 1/ln(2x)
= 2x/ln(x+x^2) - 1/ln(2x) ... Ans

I think this is the correct answer... I finally figured out how to compute the integral.

Well,

x is here just a parameter
on the other hand :
∫ 1/ln(u) du = li(u) where li(x) is the logarithmic integral (see link for definition)
therefore :

∫ 1/ln(u + x) du = ∫ 1/ln(u + x) d(u + x)

= li(u + x) + C

hope it' ll help !!

PS: and remark: when u ---> +oo, li(u) is "equivalent" to u/ln(u)
meaning : for "great" values of u : li(u) and to u/ln(u) become "close",
which translates into :
lim ( u ---> +oo ) li(u) / [ u/ln(u) ] = 1
or
lim ( u ---> +oo ) li(u) * [ ln(u)/u ] = 1