[Largedisplaystyleint_0^inftyfrac{lnleft(displaystylefrac{1+x^{4+sqrt{15vphantom{large A}}}}{1+x^{2+sqrt{3vphantom{large A}}}}
ight)}{left(1+x^2
ight)ln x}mathrm dx=frac{pi}{4}left(2+sqrt{6}sqrt{3-sqrt{5}}
ight)
]
(Largemathbf{Proof:})
(largemathbf{Method ~One:})
Assume (a,binmathbb{R}). Note that
[int_0^inftyfrac{lnleft(displaystylefrac{1+x^a}{1+x^b}
ight)}{ln x}frac{mathrm dx}{1+x^2}=int_0^inftyfrac{lnleft(displaystylefrac{1+x^a}2
ight)}{ln x}frac{mathrm dx}{1+x^2}-int_0^inftyfrac{lnleft(displaystylefrac{1+x^b}2
ight)}{ln x}frac{mathrm dx}{1+x^2}.
]
Both integrals on the right-hand side have the same shape, so we only need to evaluate one of them:
[egin{align*}
&phantom=underbrace{int_0^inftyfrac{lnleft(displaystylefrac{1+x^a}2
ight)}{ln x}frac{mathrm dx}{1+x^2}}_ ext{split the region}\
&=int_0^1frac{lnleft(displaystylefrac{1+x^a}2
ight)}{ln x}frac{mathrm dx}{1+x^2}+underbrace{int_1^inftyfrac{lnleft(displaystylefrac{1+x^a}2
ight)}{ln x}frac{mathrm dx}{1+x^2}}_{ ext{change variable} y=1/x}\
&=int_0^1frac{lnleft(displaystylefrac{1+x^a}2
ight)}{ln x}frac{mathrm dx}{1+x^2}+underbrace{int_1^0frac{lnleft(displaystylefrac{1+y^{-a}}2
ight)}{lnleft(y^{-1}
ight)}frac1{1+y^{-2}}left(-frac1{y^2}
ight)mathrm dy}_ ext{flip the bounds and simplify}\
&=int_0^1frac{lnleft(displaystylefrac{1+x^a}2
ight)}{ln x}frac{mathrm dx}{1+x^2}-underbrace{int_0^1frac{lnleft(displaystylefrac{1+y^{-a}}2
ight)}{ln y}frac{mathrm dy}{1+y^2}}_{ ext{rename} y ext{to} x}\
&=underbrace{int_0^1frac{lnleft(displaystylefrac{1+x^a}2
ight)}{ln x}frac{mathrm dx}{1+x^2}-int_0^1frac{lnleft(displaystylefrac{1+x^{-a}}2
ight)}{ln x}frac{mathrm dx}{1+x^2}}_ ext{combine logarithms}\
&=int_0^1frac{lnleft(displaystylefrac{1+x^a}{1+x^{-a}}
ight)}{ln x}frac{mathrm dx}{1+x^2}=underbrace{int_0^1frac{lnleft(displaystylefrac{x^aleft(x^{-a}+1
ight)}{1+x^{-a}}
ight)}{ln x}frac{mathrm dx}{1+x^2}}_{ ext{cancel} 1+x^{-a}}\
&=int_0^1frac{lnleft(x^a
ight)}{ln x}frac{mathrm dx}{1+x^2}=aint_0^1frac{mathrm dx}{1+x^2}=a\,Big(arctan1-arctan0Big)\&=vphantom{Bigg|^0}frac{pi\,a}4
end{align*}]
So, finally,
[int_0^inftyfrac{lnleft(displaystylefrac{1+x^a}{1+x^b}
ight)}{ln x}frac{mathrm dx}{1+x^2}=fracpi4(a-b)
]
Therefore,
[Largeoxed{displaystyleint_0^inftyfrac{lnleft(displaystylefrac{1+x^{4+sqrt{15vphantom{large A}}}}{1+x^{2+sqrt{3vphantom{large A}}}}
ight)}{left(1+x^2
ight)ln x}mathrm dx=color{Blue} {dfrac{pi}{4}(2+sqrt{15}-sqrt{3})}}
]
(largemathbf{Method ~Two:})
Let the considered integral be (I). Just to make it easier to write, let (4+sqrt{15}=a) and (2+sqrt{3}=b). Use the substitution (x= an heta) to get:
[I=int_0^{pi/2} frac{lnleft(dfrac{1+( an heta)^a}{1+( an heta)^b}
ight)}{ln an heta}\,mathrm d heta
]
Next, use the substitution ( heta=pi/2-t) to obtain:
[I=int_0^{pi/2} frac{lnleft(dfrac{1+( an t)^a}{(1+( an t)^b)( an t)^{a-b}}
ight)}{lncot t}\,mathrm dt=int_0^{pi/2} frac{-lnleft(dfrac{1+( an heta)^a}{1+( an heta)^b}
ight)+(a-b)ln an heta}{ln an heta}\,mathrm d heta
]
where I used (ln( an(pi/2- heta))=ln(cot heta)=-ln( an heta)).Add the two expressions for (I) and notice that you are left with:
[2I=int_0^{pi/2} frac{(a-b)ln an heta}{ln an heta}\,d heta=frac{pi}{2}(a-b)
]
[I=frac{pi}{4}(a-b)
]
Therefore,
[Largeoxed{displaystyleint_0^inftyfrac{lnleft(displaystylefrac{1+x^{4+sqrt{15vphantom{large A}}}}{1+x^{2+sqrt{3vphantom{large A}}}}
ight)}{left(1+x^2
ight)ln x}mathrm dx=color{Blue} {dfrac{pi}{4}(2+sqrt{15}-sqrt{3})}}
]