Let
I=int_0^1 ln(x+1)/(x^2+1) "d"x
Generally, when we have x^2+1 in the denominator we want to use the substitution x=tantheta, such that it will cancel out:
x=tantheta => "d"x =sec^2theta "d"theta
{(x=0 => theta=tan^(-1)0 = 0),(x=1 => theta = tan^(-1)1 = pi/4) :}
I = int_0^(pi/4) ln(tantheta+1)/(tan^2theta+1) sec^2theta "d"theta=int_0^(pi/4) ln(tantheta+1)/sec^2theta sec^2theta "d"theta
= int_0^(pi/4) ln(tantheta+1) "d"theta
The identity tan^2theta+1 = sec^2theta can be easily proven by writing tan^2theta as sin^2theta//cos^2theta.
Let phi = pi/4-theta:
I = int_(pi/4)^0 ln(tan(pi/4-phi)+1) (-"d"phi)=
=int_0^(pi/4) ln(tan(pi/4-phi)+1) "d"phi
Using the sum identity for tangent, which states that
tan(a-b) = (tana-tanb)/(1+tanatanb)
:. tan(pi/4 - phi) = (1-tanphi)/(1+tanphi).
I = int_0^(pi/4) ln((1-tanphi)/(1+tanphi)+1)"d"phi
= int_0^(pi/4) ln((1-tanphi+1+tanphi)/(1+tanphi)) "d"phi
= int_0^(pi/4) ln(2/(1+tanphi)) "d"phi
phi is a dummy variable, just as theta and x. We can replace it with theta and the integral would be equivalent.
:. I = int_0^(pi/4) ln(2/(1+tantheta)) "d"theta
Remember that
I = int_0^(pi/4) ln(tantheta+1) "d"theta
Hence, adding the two equations gives
2I = int_0^(pi/4) ln(1+tantheta) + ln(2/(1+tantheta)) "d"theta
=int_0^(pi/4) ln((1+tantheta)*2/(1+tantheta)) "d"theta
= int_0^(pi/4) ln2 "d"theta
2I = [thetaln2]_0^(pi//4)=(piln2)/4
Dividing by two gives us
color(blue)(I=(piln2)/8)