Question 1033411
Since {{{dtan^-1(x) = dx/(1+x^2)}}}, 

{{{int(ln(x+1)/(x^2+1), dx, 0,1) = int(ln(x+1), dtan^-1(x), 0,1)}}}.

Let {{{w = tan^-1(x)}}}.  ==> tanw  = x.  Now if x = 0, w = 0, and when x = 1, {{{w = pi/4}}}.

==> {{{int(ln(x+1)/(x^2+1), dx, 0,1) = int(ln(tanw+1), dw, 0,pi/4)}}}.

Now {{{int(ln(tanw+1), dw, 0,pi/4) = int(ln((sinw+cosw)/cosw), dw, 0,pi/4) }}}

= {{{int(ln(sqrt(2)*((1/sqrt(2))*sinw+(1/sqrt(2))*cosw)) - lncosw, dw, 0,pi/4)}}} 

={{{int(ln(sqrt(2))+ ln(sin(pi/4)*sinw+cos(pi/4)*cosw) - lncosw, dw, 0,pi/4)}}}

={{{ln(sqrt(2))*(pi/4)+ int(ln(cos(pi/4 - w)),dw,0,pi/4) - int(ln(cosw), dw, 0,pi/4)}}}.

But {{{int(ln(cos(pi/4 - w)),dw,0,pi/4) = int(ln(cosw), dw, 0,pi/4)}}} by using the substitution {{{z = pi/4 - w}}} on the left integral.

Hence {{{int(ln(x+1)/(x^2+1), dx, 0,1)  = ln(sqrt(2))*(pi/4) =highlight((pi*ln2)/8)}}}