(11)解:∫<1/2,1>dx/[x√(5x^2+4x+1)]
=∫<2,1>(-1/y^2)dy/[(1/y)√(5/y^2+4/y+1)] (令x=1/y)
=∫<1,2>dy/√(5+4y+y^2) (分子分母同乘y^2)
=∫<1,2>dy/√[1+(y+2)^2]
=∫
=∫
=∫
=∫
=(1/2)∫
=(1/2)[ln(1+sint)-ln(1-sint)]│
=(1/2)[ln(1+4/√17)-ln(1-4/√17)-ln(1+3/√10)+ln(1-3/√10)]
=(1/2)[2ln(√17+4)+2ln(√10-3)]
=ln[(√17+4)(√10-3)]。