设∫α2ln21/√etdt=π/6,求α的值.
令et-1=x2,则t=ln(x2+1),dt=[2x/(1+x2)]dx,则 ∫[1/√(et-1)]dt=∫1/x•2x/(1+x2)dx=2∫1/(1+x2)dx =2arctanx+C=2arctan√(et-1)+C 所以 ∫α2ln2(1/√et-1)dt=2arctan√(et-1)|α2ln2 故arctan√eα-1=π/4,即√eα-1=1,eα=2, 所以α=ln2.
设∫α2ln21/√etdt=π/6,求α的值.
令et-1=x2,则t=ln(x2+1),dt=[2x/(1+x2)]dx,则 ∫[1/√(et-1)]dt=∫1/x•2x/(1+x2)dx=2∫1/(1+x2)dx =2arctanx+C=2arctan√(et-1)+C 所以 ∫α2ln2(1/√et-1)dt=2arctan√(et-1)|α2ln2 故arctan√eα-1=π/4,即√eα-1=1,eα=2, 所以α=ln2.