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)#
