>>679
１
x=π-tで置換
∫[0→π]xf(sinx)dx=∫[0→π](π-t)f(sint)dt
=π∫[0→π]f(sint)dt-∫[0→π]tf(sint)dt

２
Ｉ=∫[0→π] x(sinx)dx/(1 + cos^2x)
=∫[0→π](π-t)(sint)dt/(1+cos^2t)
よって、
Ｉ=(π/2)∫[0→π] {sint/(1+cos^2t)}dt=(π/2)∫[-1→1]ds/(1+s^2)
=(π/2) arctan(s)_[-1→1]
=(π/2)(π/4+π/4)
　=(π/2)^2

３
∫[0→∞] dx/(x^2+4x+1)
=∫[0→∞] dx/{(x+2-√3)(x+2+√3))
=∫[0→∞](1/(2√3)){1/(x+2-√3) -1/(x+2+√3)}dx
=1/(2√3) log|(x+2-√3)/(x+2+√3)|_[0→∞]
　lim[x→∞] log|(x+2-√3)/(x+2+√3)|
　=lim[x→∞] log|(1+(2-√3)/x)/(1+(2+√3)/x)|
　→log1=0
だから、与式は、
-1/(2√3)log{(2-√3)/(2+√3)}