19/02/05 10:59:13.23 xI3EwwZt.net
>>692
(上)
(2xx+1)/(xx+x+1) = 2 - (2x+1)/(xx+x+1),
∫(2xx+1)/(xx+x+1) dx = 2x - log(xx+x+1) +c,
(下)
X = πxx, Y = πyy とおくと
dX = 2πx dx, dY = 2πy dy
∬ xy sin(π(xx+yy)) dx dy
= (1/2π)^2 ∬ sin(X+Y)) dX dY
= (1/2π)^2 ∫{ -cos(X+Y)} dY
= -(1/2π)^2 sin(X+Y)
= -(1/2π)^2 sin(π(xx+yy)),