lf ∫∞0x2dx(x2+a2)(x2+b2)(x2+c2)= π2(a+b)(b+c)(c+a) then ∫∞0dx(x2+4)(x2+9)=
∫∞0x2dx(x2+a2)(x2+b2)(x2+c2)=π2(a+b)(b+c)(c+a)
∫∞0x2dx(x2+22)(x2+32)(x2+0)=π2(5)(3)(2)
=π60.