∫1/[x√(1+x²)]*dx
設(shè)x=tant,dx=1/cos²t*dt
1+tan²t=1/cos²t
√(1+tan²t)=1/cost
原式=∫cost/sint*cost*1/cos²t*dt
=∫1/sint*dt
=∫csctdt
=ln|tan(t/2)|+C
tan(t/2)=(1-cost)/sint=(1/cost-1)/(sint/cost)=[√(1+tan²t)-1]/tant
原式=ln|[√(1+tan²t)-1]/tant|+C
=ln|[√(1+x²)-1]/x|+C
=ln{[√(x²+1)-1]/|x|}+C
∫√(x²-4)/x*dx
當(dāng)x>0時(shí),設(shè)x=2/cost
dx=2sint/cos²t*dt
tan²t=1/cos²t-1
tant=√(1/cos²t-1)
cost=2/x
t=arccos(2/x)
原式=∫2tant*cost/2*2sint/cos²t*dt
=2∫tan²t*dt
=2∫(1/cos²t-1)dt
=2∫1/cos²t*dt-2∫dt
=2tant-2t+C
=2√(1/cos²t-1)-2t+C
=√(4/cos²t-4)-2t+C
=√(x²-4)-2arccos(2/x)+C
當(dāng)x