回答:
#sqrt(1 + x ^ 2)-1 / 2 ln(abs(sqrt(1 + x ^ 2)+ 1))+ 1/2 ln(abs(sqrt(1 + x ^ 2)-1))+ C#
説明:
つかいます #u ^ 2 = 1 + x ^ 2#, #x = sqrt(u ^ 2-1)#
#2u(du)/(dx)= 2x#, #dx =(udu)/ x#
#intsqrt(1 + x ^ 2)/ xdx = int(usqrt(1 + x ^ 2))/ x ^ 2du#
#intu ^ 2 /(u ^ 2-1)du = int1 + 1 /(u ^ 2-1)du#
#1 /(u ^ 2-1)= 1 /((u + 1)(u-1))= A /(u + 1)+ B /(u-1)#
#1 = A(u-1)+ B(u + 1)#
#u = 1#
#1 = 2B#, #B = 1/2#
#u = -1#
#1 = -2A#, #A = -1 / 2#
#int1-1 /(2(u + 1))+ 1 /(2(u-1))du = u-1 / 2ln(abs(u + 1))+ 1 / 2ln(abs(u-1) )+ C#
パッティング #u = sqrt(1 + x ^ 2)# 戻って与える:
#sqrt(1 + x ^ 2)-1 / 2 ln(abs(sqrt(1 + x ^ 2)+ 1))+ 1/2 ln(abs(sqrt(1 + x ^ 2)-1))+ C#